Arithmetic Progressions Fastracl« Revision » Some numbers arranged in a definite order according to a definite rule, are sald to form a sequence, where each umber Is called a term. > A sequence whose terms follow a certain rule, Is called a progression, » Asequence in which each term differs from its preceding term by same constant except the first term is called an Arithmetic Progression (AP). This constant is called the common difference (d) of the AP. It can be positive, negative or zero. > The general form of an AP Is: 9,0 + d,a + 2d, .. where ais the first term, > IF 0, b,c are three terms in AP, then b-a = c- 6, le, 2b=a+e > AN AP with finite number of terms isa finite AP and which does not have finite number of terms is an infinite AP. Infinite AP's do not have a last term. > In an AP, if we add, subtract, multiply or divide each term by the same non-zero number, then the resulting sequence would always be an AP. BOOSTER 1. Three terms in AP should be taken as: a — 4, a, avd. 2. Four terms in AP should be taken as:a-3d,a—d, o+da+ 3d. 3. Five terms in AP should be taken os: a—2d,a—d, g.0+d,0+2d. > General Term of an AP: If the first term oF an AP is o, common difference is d and its last term Is , then its nth or general term is given by T,= 0,= [= a+ (n—1)d. > IF nth term oF an AP is a,, then the common difference is determined by d= 0,~0,_ 1 > nth Term From the End of an AP: Ifthe First term of an AP isa, its common difference is dand its last term Is , then nth term from the end = I~ (n= 1) d. ‘and pth term from the end = (n~ p+ 1)th term from the beginning. » Sum of m Terms of an AP: If 5, is the sum of n terms oF a zlo+) anap,then S,=SI2a+(a-thd] oF 5 where ais the First term and lis the last term, wledge BSOSTER 1. If sum of n terms (S,) of an AP is given, then nth term (o,) of an AP can be determined by 0n=5,—Sq-1 and common difference d= 'S)—2Sy-1 + Sy 2. Arithmetic Mean (AM) between Two Numbers: If Ais the AM between a and b, then In Ont , axb R z xB Practice Exercise -@ Multiple choice Questions QL. Sthterm oftthe sequence, whose mth term is 4n « 2,is: 2.20 b22 cB 4.23 Q2 The common difference of the AP 1 1-p 1-29 = ae a ts: pop! Pp al b.vp c-1 d-Vp Q3. If k + 2, 4k —6 and 3k — 2 are three consecutive terms of an AP, then the value of kis: (cese 2023) aa b-3 c-4 a4 ‘The first four terms of an AP, whose first term is —2 and the common difference is ~2, are: INCERT EXEMPLAR) b.-2.4,-8.16 d.-2.-4,-8,46 Qa 2-20.24 © -2-4-6.-8 The next term of AP: /7, 28, /63 is: (case 2023) a ¥70 b. ¥B0 <7 a. Viz 6. The next two terms of an AP sequence 5, 8, 1: are: 2.13.16 Qs. bw cid 1215Q7. 98. Q9. quo. qu. giz gus. Qu. us. Que. Qu. Que. qus. Q20. In an AP, if a= 5,d~3 and n~ 10, then the value of Og 18: al ba 34 0.36 ‘The number of terms of an AP, having first term 5, common difference 3 and last term 74, is: a2 b 24 25 26 The nth term of the AP: 0,3a,5a,.. is: (cBSE2020] ano b. (2n-1)o no 4. 2na Which term of the AP: 21, 42, 63, 84,...is 2107 [NCERTEXEMPLAR] a. 3th b. 10th c.Tth 4. 12th ‘The 21st term of the AP, whose first two terms are 3 and 4,is: [NCERTEXEMPLAR] a bay 14a 4103 Ifthe common difference of an AP Is 5, then what is 045-043? INCERTEXEMPLAR] as b 20 25 4.30 ‘The 7th term from the end of the AP: 17,141: a-18 c-2% 4-20 If the 2nd term of an AP is 13 and the Sth term is 25, what is its 7th term? [NCERT EXEMPLAR} a.30 baa oe 38 ‘Two AP's have the same common difference. The first term of one of these is-1 and that of the other is -B. Then the difference between thelr 4th term [cese sop 2023-24) a9 If the sum of the first terms of an AP be 3n? +1 and its common difference is 6, then its first term al ba <7 i [cese2023) a2 ba cl a4 Ifthe sum of n terms of an AP is S, = 2n’” » 3, then ‘common difference of an AP is: a3 ba c2 d-2 ‘The sum of first 7 terms of an AP: 2,4, 6,8, 10,..is: a.52 b. 54 56 458 In an AP, if a= 1,0, = 20 and S, = 399, then nis equal: 3.38 b39 < 40 dal The number of terms of an AP: 64, 60, 56,..., whose sum is 544, aie bia C1415 A718 Assertion & Reason type Questions y Directions (Q. Nos. 21-27): in the following questions, statement of Assertion (A) is followed by a statement of Reason (R). Choose the correct option ‘2. Both Assertion (A) and Reason (A) are true and Reason (R) is the correct explanation of Assertion (A) ga. Qe, Q23. Q 24. gas. Q 26. Qa7. 926. Q29. Q30. aL .. Qs2. b. Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A) c. Assertion (A) Is true but Reason (R) is false d. Assertion (A) is false but Reason (R) is true Assertion (A): is in Arithmetic Progression. Reason (R): The terms of an Arithmetic Progression cannot have both positive and negative rational numbers. [cose sop 2023-24] Assertion (A): The nth term of the sequence ~8,~4,0,4,.ui8 4-12. Reason (R): The nth term of an AP is determined by T,=a+(n—1)d. Assertion (A): The common difference of an AP in which ay9— dye = 20 is 5 Reason (R): The nth term of the sequence V2.V4,NI8,..is Van Assertion (A): a, b, ¢ are in AP if and only if 2b=a+c. Reason (R): The sum of first n odd natural numbers isn’, [cose 2023] Assertion (A): The sum of first 20 even natural numbers divisible by 5 is 2110. Reason (R): The sum of n terms of an APis given by 5, = 2 (a0, where (isthe last term ofan AP Assertion (A): The sum of the series with the mth term T, = 7 —3n Is -255, when number of terms is n=15. Reason (R): The sum of AP series is determined by 1 5, =$2a+(n—1)4) Assertion (A): If sum of first 1 terms of an AP is 5, = 6n" — 2n, then nth term of an AP is 120-8. Reason (R): Suppose S, be the sum of n terms of an AAP, then nth term of an APis T, = S, -~ Sp Fill in the Blanks type Questions w If the common difference hen each term of the AP will be same as the first term of the AP, The value of k for which K? + 4k + 8, 2K + 3k~ 6, 3K + 4k +4 are three consecutive terms of an AP, is . INERT EXEMPLAR] If7 times the 7th term of an AP is equal to 11 times its 14th term, then its 18th term will be term of the AP: 24, 21, 18, 15, negative term. The sum of 10 terms of an AP: — 8, - 6, is.. isthe first933, Q34. True/False Type Questions y If we multiply each term of an AP by 2, then the resulting sequence is an AP. In an AP, ifa=1,a,=20 and n= 38, then sum of first 38 terms is 499. Solu’ (0) Given, a,= 4042 Put n=5. we get 05" 4(5) +2 =20+2=22 (o) Common difference =p) pp (@) Given terms of an AP are k+2,4k~6 and 3k-2 Therefore, common difference of each term should be equal. So. (4k ~6) ~(k +2) = (3k 2) (4k-6) > 3k-8=-k+4 = ko > kod (0) Let o be the first term and d be the common difference of an AP. Then 2 and d=-2 First four terms of an AP are a.a+da+2d.a+3d, Here on-2 asd--2-2-~4 0+ 2d=-2+2(-2)=-2-4=-6 and a+ 3d=-2+3(-2)=-2-6 Hence, first four terms of an AP are -2, +4, -6, -8. () Given ap/7.128. 63. Here, a, =V7.0, «V2 = Van 7 = 7 and 03 = N63 =V9x7 = W7 wi-V7=V7 Common difference (d) =0,~0, = So. next term ie. 4th term of APis a, 20,+(4-l)d 5 ogo VF +i «ti. oN aia TD (b) Given AP: 5, 8.7 Here. 0, =5, 0,=8 and o,=1 Common difference (d)= 0, -0)=8-5=3 So. next two terms Le, 4th and Sth terms of AP are aeq+ (4-1) do5+3x3=5+9014 and 0; = 0)+ (5-1) d=5+4x3=5+12=17 0,2 0+ (n= 1)) Q35. 936. 937. If the first term of an AP is -5 and the common difference is 2,then the sum of the first 6 terms is 5. 20 terms of AP: 18, 16, 14,... should be taken so that their sum is zero. If sum of the first 1 terms of an AP is given by S,=3n? +4, then its nth term is 6n—3. ns 7. (0) Given. a= 5.d=3 and n= 10 0,=0+(n—1)d og = 5+ (10-1) «32 5427-32 8. (b) Given. o=5.d=3 and last term (0) = 74 n= 24 So, the number of terms is 24 9. (b) Given. first term (A) =o, common difference (D) = 30-0=20 +s mth term of AP, 0, A+(n=1)0 0, 0+ (n=1)2a" 0+ 2an-2o =2on-a=(2n-1o 10. (b) Given AP sequence is 21.42. 63, 84 Here, first term. a= 21 Common difference. d= 42-21=21 and last term, (= 210 Then. nth term of an AP sequence is given by Ty=l=o4(n—Id 210 = 21+ (n-1)21 = -2=21+2In-21 = dn=20 = n=10 Hence, 10th term of an AP is 210. WL (b) Given, =—3and a4 Here, comman difference. d= 0,~0,=4~(-3)=7 and first term, a=-3 Then nth term of an AP is given by Tp=0+(n=Nd Therefore, 21st term of an AP is Tyo -3+ (21-107) 34+20x 3+140=137 12. (c) Let o be the first term and d be the common difference of the given AP. Given. d=5 TRICK nth term of an AP is given by 0,=0 + (0=1)d + (18 -N)d- (a+ (13 -1)d) Vd~12d= Sd= 5x5 = 25 Op ~ 313. (b) Given AP is: 17,14, 11 Here, first term, a= 17, Common difference. d= 14-17 =~3 and last term, [=— 40 TR!CK nth term from the end of an AP is aye. 7th term fram the end of an AP ==40-(7=1)(-3) =-40-6 (-3)=—40 + 18 =-22 14, (b) Let @ and d be the first term and common difference of an AP respectively Then, nth term of an AP is, =40 T,=0+(a=1d itis given that heB and T.=25 a+(2-Id=B > o+d=13 a anda (5— id= 25 = a+ 4d=25 (2) Subtract eq. (1) From eq, (2). we get (0+ 4d) -(0+d)=25-13 = 3-2 > Put d= 4 in eq, (1). we get a+4=13 > 0-9 Teh term of an AP is Ty=94(7-1]4=9+6%4=9424=33 15. (0) Let first term of first AP and second AP be a and 0" respectively. The same common difference of both APisd Given that. a» =1 and o' = ~8 4th term of first AP. a, = 0+ (4—I)d=-1+3d and 4th term of second AP, o'y = o + (4— Iho =-8430 So. difference between 4th terms af both AP oe 0'a =(1+3d)- (8+ 30) =-1+ 34+ 8-3d=7 16. (d) Given, sum of first‘ terms af an AP, S,<30t+n Then, nth term of an AP is determined by 0, Sp 5-12 (30? +) - 3(n- 1)? -(n-1) = an? +n-3(n? +1=-2n)—(n-1) =30? +n-3n?-3+6n-n+1 =6n-2 First term of an APIs o)= 61-2 2 6-2-4 (put n= 1) 7. (b) Given. §,,= 2n? + 3. Then, nth term of an AP is determined by q* Sy~ Spor (2r? +3) ~(2(-1)? +3) =(2n? +3)-(2n? 4+ 5)=4n-2 Now. the common difference of an AP is given by d= 0,~ 0,19 4n~2~(4{a=1)~2) 24n-2-4nehe2=4 18. (c) Given sequence of an AP is 2. 4, 6, 8. 10. ..... Here. first term, a= 2, 9. a. Common difference, d=4—2=2 The sum of first n terms af an AP Is given by 5 =§ a+ (n-1)d) ‘The sum of first 7 terms of an AP is 7 7 5 = Ztax2+(7-12) = Px2i2+(7-) T246]=7x8=56 (a) Given, o=1. 0,= 20 and 5, = 399, ‘The sum of n terms of an AP is given by 5, = $l20+(n—Hd) 4 389 = F(21+(0-N)d] = 798 = 2n+ n(n= 1d a) Also. g,=20 a+ (n= 1d > 1+ (n=1 = (p= )d=19 Put (n= 1)d = 19 in eq, (I), we get 798 = 2+ 190 => 79882In > 1838 (b) Given AP is 64, 60, 56, Here, 0 64, d= 60-64 =~4 Let nbe the number of terms in the given AP. Then, 5, = 544 Fl20+(n—Nd}= 544 = Fl2x64.+(n—1)(-4)}= 544 = §x2f64-2In-n)] =544 = (65 ~ 2n) = WP 66+ 544 = n?—33n+272=0 = n-(17+16)n+272=0 (by splitting the middle term) = m-1n-16n+ 27250 = afn=17)=16(n-17 = (n=16)(n-17)=0 = n-16=Oorn-17 = n=16 or n=17 (c) Assertion (A): Given sequence: -5. Here, Difference af two cansecutive terms:Since, the difference of two consecutive terms is 5 constant Le. 8, “ 2 Therefore, given sequence is an AP. So, Assertion (A) is true. Reason (R): The terms of an AP. can have both positive and negative rational numbers. So, Reason (R) is false. Hence. Assertion (is true but Reason (R) is false. 2. (a) Assertion (A): Given sequence Is - 8.-4. 0. 4. 0y- 9)=-4—-(-8) =4. 0-0; 0-4) 4, Q-0;=4-0=4 Here. we see that difference of two consecutive terms: is same constant. So, given sequence is an AP. First term a2 -8 and camman difference, d= 4 TR!CK nth term of an AP is Ty=0+ (nd Ta=-8 + (n=) (4) 2B 44n—4= 4-12 So. Assertion (A) is true. Reason (R): Its also true that nth term of an AP Is determined by T,= 0+ (n- Id Hence, both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A). (0) Assertion (A): Let o and d be the first term and common difference of an AP. Then nth term of an APis a,-0+(n=I)d Given, = O45 =20 (0+ (20- Nd) - (0+ (16-1) 19d~15d= 20 = 4d=20 = des So, Assertion (A) is true. Reason (R): Given sequence is V2.v4.018, or V2.2N2.3V2, Here a= ¥2.d=2V2-J2=3V2-2W2=v2 T,=0+(0-Nd Ty =N2 +(n-1W2 = J2n So, Reason (R) is also true. Hence. both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A). (b) Assertion (A): If part: Given 0, b.carein AP. Then b-a=c-b = b+b=0-c > 2b= 25. Only part: Given, 2b=0+¢ > b+b=arc = b-o=c-b > 01-0,=0;- 0 (let o,=a.0)= band =<) Since. each term differs from Its preceding term are equal, The sequence a, 0», 03 oF 0, b. care in AP. Therefore, a,b, care in AP if and only If 2b= a+ So. Assertion (A) is true. Here, first term (a) =1 and common difference (d) = Since, the difference between each consecutive terms is constant ie, 2. So, the sequence forms an AP. ‘Sum of first n terms of an AP, (20+ (n-1)d) eB (2x14 (x2) x2(l+n-=n-n=n? So, Reason (R) is true Hence, both Assertion (A) and Reason (R) are true but Reason (R) Is not the correct explanation of Assertion (s). (4) Assertion (A): Even natural numbers divisible by 5 are 10, 20. 30. 40. They form an AP with first term, a= 10 and common difference, d= 20 ~ 10 = 30-20 = 10 [20+(n=1)o] 20 2eex10+ (20-10) = 10(20 +190) = 2100 So, Assertion (A) is false. Reason (R): tis true that the sum of n terms of an 0 APIs 5, =S(a+0), So, Reason (R) is true. Hence, Assertion (A) is false but Reason (R) is true. (a) Assertion (A): Given, nth term is T,=7—3n. Here. =3()=4 Ty=7-3(2)=1 Ty=7-3(3)--2 Here, sequence is 41.2 Now. 1-4=-3, -2-12-3 Here, difference of two consecutive terms fs same. So. it Is an AP. Here, first term a =4 and common difference d: ‘The sum of 15 terms of an AP is Sis 5, n Beaxa+(5— 1-3) | $= Fla +(0-Ne) 5 5 Bie~4a} = 2x(-34)=-255 So, Assertion (A) is true.za. 28. 2. 30. aL Reason (A): Its also true that sum of n terms of an AP is determined by 5, = 2(2a+(n-I}d), 2 Hence, both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A), (0 Assertion (A): Given. 5, = 677 -2n. TRICK ith term of an AR whose sum is S,, 15 Using formula, T= Sy~ Sy-1= (Gr? = 2n) = [6(n=1)?-2(n- 1)] Gr? - 2n- (6(n? + 1- 2n)- 2042) = 61? -2n-[6r?~ Man + 8) =-2n+ ¥in-8=12n-8 So, Assertion (A) is true Reason (R): Its not true that Ty=Sn-1- Sh ‘Thus, the correct relation is Ta Sn~ Saat Hence, Assertion (A) Is true but Reason (R) Is fase. Zera Given. three consecutive terms af an AP are I? + 4k +8, 2ie + 3k+ 6 and 3? + 4k + 4. ‘Therefore, common difference of each term af an AP Is equal. Here, = H2+4k+B, 0,= 212 +346 and = 3 +k +4 2 0)=05~ 0 = (2 + 3k + 6)= (12 +k + 8) (312 + i+ 4) ~ (217 + 3k + 6) = Pak=2a ek? = 2k=0k=0 Let abe thefirst term and dbe the common difference ‘of an AP. Then according to the given contin. TTy=MTy 7 (0+ (7d) =N fo (N= N)d] fe T= 0+ (9=Nd) = Mo+Sd)=11[0+ 10) = Ta+42d=Mlo+ Md = 40+68 = ov d=0 0) Now. the 18th term of an AP is Te=0+(18-1)d sa+tl {from eq, (1) Letnth term of the AP: 24,21, 18,15, ..... be the first negative term. Here. first term (a) = 24 and common difference (d) = 21-24 =-3 4 a,=a+(n-0d (o+(n-1)d}<0 = 24+ (n-9-3))<0 = (24-3n+3)<0 = 27-3n<0 > 3n>27 > n>9 ns 10 So. 10th term is the first negative term. {EOMMONJ ERROR Sometimes students take the value of n © 9 instead of taken the value of 0 = 10. 32. Given AP I5-B.-6,-4 Here, first term, a=—8 and common difference, d=- 6 -(~8)=2 Then, sum of an AP is 5, = 5(2a+(n-Nd) a) So= l2x(-8)+ 00-2) 5[-16 +18) =5x2=10 True Given, a= 1, q, 220 and n= 38 eH Q 5, = 5(e+0,) Sq= +20) = Syg= 1921 = S99 399 Hence, glven statement s false 35. Let cand de the first term and common difference respectively. Then ae-5 and d=2 ‘The sum of frst m terms of an AP is Sq = G(20+(n-d} 6, Fl2x(-5)+(6-2) =3{-10+10)=0 Hence. given statement is false. 36. Given, APIs 18,16. 14. Here. a= 18. d= 16-18 =-2 Now. sum of 20 terms of an AP is (2x18+ (20-1)(-2)) S20 a ang = 1036-38) = 10 (-2)=-20 Hence, given statement Is false 37. Given, S,=3n? +4 ith term. = 5)— Sj ar? +4 -[3(n-1)? +4) i? +4 (a(n? + 1-2n) + 4) =3n? 44-307 +6n-7 3 Hence, given statement is true (20+ (ow“@ Case Study Based ouestions Case Study1 India is competitive manufacturing location due to the low cost of manpower and strong technical and engineering capabilities contributing to higher quality production. The production of air conditioner in a factory increases uniformly by a fixed number every year. It produced 12000 sets in 3rd year and 20400 in 10th year. Based on the above information, solve the following questions: QL. Find the production during first year. aseg0 6 200 «¢ 2400 © d. 8000 Q2. Find the production during 8th year. a. 9600 , 18000 < 8000 14400, Q3. Find the production during first five years. a. 55400 b. S0B00 < 60000 . 62600 Q4. In which year, the production is 30000? als b16 cv ae QS. Find the difference of the production during 7th year and Sth year. 22400 b 1200 9600 d.4000 Solutions * 1. Given that, the production of alr conditioner in a factory increases uniformly by a fixed number every var. ie. production of air conditioner in every year form an AP, Let the First term and common difference of this AP be ‘a’ and’d' respectively According to the question, the factory produced 12000 alr conditioner in 3rd year. TRICK. ith term ofan AP is T, =a. (n—4) d where, a and d are first term and comman difference respectively. ie Ty=0+(3-)d = 120000424 () ‘and 20400 air conditioner in 10th year. ie Te=o+ (10-1 = 20400 =0+9d (2) Now subtracting eq, (1) from eq, (2). we get (0+ 9d) - (a+ 2d) = 20400 - 12000 = 7d~ 8400 a £406. = 1200 - Put the value of din eq. (1). we get 12000 =0+2% 1200 > 12000 =0+ 2400 a= 12000 - 2400 9600 Hence. the production during first year Is 9600. So. option (a) is correct. 2. The production during 8th year Is. Ta=0+(8=1)0 = 9600 +7 «1200 =9600 +8400 = 18000 So, option (b) is correct. TRICK The sum of n terms of an APis S, = $0 +(n=1)d) The production during first five years is 5, 3(2*9600+(5-m200] 39200+4 1200) 39200 +4800) 5 =2x24000 2 =5x12000=60000 So, option (c) is correct. 4. Let nth year. the production is T,= 30000 > 0+(n=1)d=30000 = 9600 + (n-1) 1200 = 30000 20400 1200 nei7+1=18 Hence. in 18th year. the production Is 30000, 50, option (d) is correct. 5. Now, the production during 7th year is (T,= a+ (n=1d) T,=0+(7-)d [+ T= 0+ (n-1)d) = 9600 + 6 x 1200 9600 + 7200 = 16800 and the production during 5th year is o+(5=1)d S600 + 4 x 1200 = 9600 + 4800 = 14400 TsThe difference of the production during 7th year and 5th year = 16800 - 14400 = 2400 So. option (a) is correct. Case Study 2 Your younger sister wants to buy an electric carand plans to take loan from a bank for her electric car. She repays her total loan of & 321600 by paying every month starting with the first instalment of % 2000 and it increases the instalment by % 200 every month. Based on the above information, solve the following questions QL Find the list of the instalment formed by the given statement. a 200,180,160... b. 200.2200, 2400. 200, 2400, 2600... d. 2300, 2600, 2900. ‘The amount paid by her in 25th instalment is: a%6800 b 3500 «4800 4 t6600 Q3. Find the difference of the amount in 4th and 6th istalment paid by younger sister. Q2 at200 be400 cz600 4 ta00 Q4. In how many instalment, she clear her total bank loan? a 1582 b 1580 ce 1899. 1600. QS. Find the sum of the first seven instalments. ae14000 b.e13600 «210400 4 ¢12600 Solutions 1 It can be observed that these instalments are in AP having first term (instalment) as @ 2000 and common difference (increase instalment) as 2200. Here, 0 = 2000 and d=200 Therefore Uist of an APIs 0, 0+ d. 0+ 2d, le. 2000, 2000 +200, 2000 + 2 « 200. le, 2000, 2200, 2400, So, option (b) Is correct. 2 Itcan be observed that these instalments are In an AP having first term (Instalment) as € 2000 and common difference (increase instalment) as & 200. Here, 02000 and #200 TRICK ath tem afan APIs, Ty= + (n= 1) vihere a and dare fst term ond common difference respectively. The amount paid by her in 25th instalment is Tys= 04 (25-1) = 2000 + 24 x 200 = 2000 + 4800 = € 6800 So, option (a) Is correct. 3, Lot aand dbe the first term and common difference of an AP, Then, — a=0+ (4-1 t =0+3d Similarly. o¢ = 0+ Sd Required difference = a5 0, = (0+ 50) -(0+3d)=20 =2. 200-2400 ‘50, option (b) is correct. 4, Letin n instalments, she clear her loan Given, T,= 321600 Ta= 0+ (n= 1d 321600 = 2000 + (n—1)200 = 319600 = (n= 1)200 = 1598=n-1 = n=1599 So. in 1599 instalments, she clear her bank loan So, option (¢) is correct 5, Here, a= 1200, d= 200 9,=0+(n=1)d) The sum of first seven instalments is 5 22 x1200+(7-1)200) (2400+ 1200) (3600) = 71800 = 12600 So, option (d) is correct Case Study 3 In an examination hall, the examiner makes students sit in such a way that no students can cheat from other student and make no student sit uncomfortably. So, the teacher decides to mark the numbers on each chair from 1, 2,3, 0. ‘There are 25 students and each student is seated at alternate position in examination room such that the sequence formed is 1, 3, 5,Based on the above information, solve the following questions: QL What type of sequence is formed, to follow the seating arrangement of students in the examination room? Find the seat number of the last student in the examination room. Find the seat number of 10th vacant seat in the examination room. Q2 Q3. Qn Qa But he wants to achieve a target of 3900 push-ups in Lhour for which he practices regularly. With each day of practice, he is able to make 5 more push- ups in one hour as compared to the previous day. If on first day of practice he makes 3000 push-ups and continues to practice regularly till his target is achieved. Based on the above information, solve the following questions: {Ic8S€2022 Term] Form an AP representing the number of push-ups, per day and hence find the minimum number of days he needs to practice before the day his goal is accomplished. Find the total number of push-ups performed by Nitesh up to the day his goal is achieved. Solutions * 1. Given, seating arrangement of students in the ‘examination room is 1.3.5, Here, 0)=1.0)=3.0)=5. Now. @,-9, oy- 0, 25-352 Here, common difference is same. so given sequence is the type of Arithmetic progression 2 Gren, o=1,du3-1e2andn=25 Ty= 0+ (n=I)d There are 25 students. Tyg2 1+ (25-1)221+26%2=49 Hence, last student will sit on the 49th seat number. 3. The sequence of vacant seats are as follows, 2.4.6. 4B, Here, o=2.d= 6-4-2 The 10th vacant seat wil be T= 0+ (10-16 249%202418=20 Hence, the 10th vacant seat number Is 20, Case Study 4 Push-ups are a fast and effective exercise for building strength. These are helpful in almost all sports including athletics. While the push-up primarily targets the muscles of the chest, arms and shoulders, support required from other muscles helps in toning up the whole body. Pos e@osy Nitesh wants to participate in the push-up challenge. He can currently make 3000 push-ups in one hour. (Th a+ (n= Nd} Solutions. - In first day, Nitesh makes 3000 push-ups and he is increasing 5 push-ups each day. Therefore first term, a= 3000 ‘and common difference, d=5 AP sequenceis a.0+d.042d.0+36.... 3000, 3000 + 5,300 +25, 3000 +35, ‘or 3000, 3005. 3010, 3016. itis glven that o, = 3900 0, =04 (nd 3900 = 3000 +(n=1)5 > (n-1)5 =300 = n-1=180 = n=181 Hence, minimum number of days he needs to practice before the day his goal accomplished is 181 ‘The total number of push-ups performed by Nitesh to the day his goal achieved is 2fa+a,] 512+%n, 181 2 - 2.6800 2 {3000+ 3900) = 18) x 3450 = 624450 Case Study 5 ‘The school auditorium was to be constructed to accommodate at least 1500 people. The chairs are to be placed in concentric circular arrangement in such a way that each succeeding circular row has 10 seats more than the previous one.Based on the above information, solve the following questions. [case sqp 2022-23] QL. If the first circular row has 30 seats, how many seats will be there in the 10th row? 2 need to be there? Or If 1500 seats are to be arranged in the auditorium, how many seats are still left to be put after 10th row? If there were 17 rows in the auditorium, how many seats will be there in the middle row? Q3. Solutions - 1. Since, each raw is increasing by 10 seats. so itis an AP with first term a=30 and common difference d=10. Ser OSE TO MoH p = [-o,=0+(n-1) 3049.10.20 2 TRICK The sum of n terms of an AP is S, F120 +(n—1)a] 1500 = 2230+ (n-1)10) = 3000 = 50n + 10r? = 1m +5n-300= = fs 20n—15n-300 <0 (by splitting the middle term) > nla+20) -15(n+ 20) =0 = (0+ 20) (n-15) =0 = n=-20,15 Rejecting the negative value, so we consider n=15 So, 15 rows need to be there. a QL Q2 Q3. So = Baxs0+(0 = 110) = 5(60 +90) = 750. So, the number of seats still required to be put are 1500 - 750 = 750 Given, number of rows = 17 o520+(9-1)d =0+8d=30+8x10=N0seats So, 0 seats will be there in the middle row: For 1500 seats in the auditorium, how many rows Case Study 6 Manpreet Kaur is the national record holder for ‘women in the shot-put discipline. Her throw of 18.86 m at the Asian Grand Prix in 2017 is the biggest distance for an Indian female athlete. Keeping her asa role model, Sanjitha is determined to eam gold in Olympics one day. Initially her throw reached 7.56 m only. Being an athlete in school, she regularly practiced both in the mornings and in the evenings and was able to improve the distance by 9 cm every week During the special camp for 15 days, she started with 40 throws and every day kept increasing the number of throws by 12.o achieve this remarkable progress, Based on the above information, solve the following questions. [CBSE S0P2023-24) How many throws Sanjitha practiced on 11th day of the camp? What would be Sanjitha’s throw distance at the end of 6 weeks? or When will she be able to achieve a throw of 11.16 m? How many throws did she do during the entire camp of 15 days? or Number of seats already put up to the 10th row =S0 COMMON] ERROR . in haste, some of the students consider negative value i of n. So, please be careful of such type of problem: Solutions * List of the throws during the camp: 40, (40 + 12), (40 +212), (40 +3 x 12). ie. 40, 52.64, 76 Here the difference of two consecutive terms shows constant Le, 12Qe. So, the above sequence forms an AP. Let o'and lo'be the first term and comman difference of an AP respectively, a3. o= 40 and d= 52-40 =12 (On Tith day le, n=T1 04 0, = 0+(n=I)d a= 40 + (1-1) (12) = 40-+10 x 12 = 404120 = 160 As So. 160 throws Sanitha practiced on Nth day of the camp Q6. 2. List of improve distance in throws by Sanjtha is: 756 m, 756 m +9 em, 756 m +2x9 em, Le, 756 m, 756 + 0.09 m. 756 +2 «0.09 m. ie. 756, (756 + 0.09), (756 + 018). Le, 756. 765.774, Here the difference of two consecutive terms shows constant (e., 0.09. So. the above sequence forms an AP. Lot o’and be the first term and common difference of an AP respectively. Then, 0 =756 and d= 765 -756 = 0.09 On 6th week Le. n= 6, Q7. 6. Q9. que. a,=0+ (n=I)d au 9¢= 756 + (6~1)x0.09 756 + 0.45 = 601 gz. ‘So. Sanjitha’s throw distance at the end of 6 weeks is 6.01m. Or Q3. Let she will be able to achleve a throw of 1116 min a weeks. a4. Here, first term (0) = 756, common difference ()= 0.09 and o, » 1116. 9,¢0+(n=1)d TG = 7.56 + (1-1) « (0.09) 009 (n-1)=36 > n-1-38 0: Qs. Qo. 40 n=4041=41 So, Sanjitha’s will be able to throw T116 min 41 weeks, 3B. From Ust of part (I), 40, 52, 64,76 Q7. rT Here, o=40,d=52—40=12 and n=15 (20+(n-1a) a9. Sig = Bla 40. (15-112) =15(40+ 4x6) Q10. 15 (40 + 84) = 15 x 124 = 1860 qu. So, 1860 throws she do during the entire camp of 15 days. . a Very Short Answer type Questions W QL. Write the common difference of the AP: Qs. v3, V2, V27, V48,.... Qu. INCERT EXEMPLAR; CASE 2019] Suppose a=3,d terms exist in an AP. What is the common difference of an AP in which 84? [cose 2017} In an AP, if the common difference (d) is~ 4 and the seventh term (a;) is 4, then find the first term. [case 2018) in the AP: 2,[=~ 11, find the number of 4-0; Find the 1 48, 15>,13, number 47. of terms [NCERT EXERCISE; CBSE2019} How many two digit numbers are divisible by 3? INcERT EXERCISE; COS 2019} Find the 9th term from the end (towards the first term) of the AP: 5,9, 13,... 185. [cose 2016] Ifa=2and d=3,then find the sum of first 10 terms of an AP. How many terms of AP: 18, 16, 14, ... should be taken so that their sum is zero? If nth term of an AP is (2n + 3), what is the sum of its first five terms? Q Short Answer tTyse-! Questions y id whether ~150 is a term of the AP: 17,12,T, Bye « [wimpy Which term of the AP: 8, 14, 20, 26, L be 72 more than its 41st term? [ces€ 2017] In an AP, if the sum of third and seventh terms is zero, find its Sth term. {[¢85€ 2022 Term-t] IF 7 times the seventh term of the AP is equal to 5 times the fifth term, then find the value of its 12th term. [cose 2022 Term-t) ine the AP whose third termis 5 and seventh 9. [c6SE 2022 Term-tl] For what value of n, are the nth terms of two AP's 63, 65, 67,...and 3, 10, 17,... equal? INCERT EXERCISE; COSE2017] Determine the AP whose third term is 16 and 7th. term exceeds the Sth term by 12. INCERT EXERCISE; C8SE2019} Find the middle term of the AP: 6,13, 20,....216. [cosé2015) Find the sum of first 30 terms of AP: —30,-24,-18, [c0S€ 2022 Term] In an AP, if S, =n(4n-+1), then find the AP. [c9S€2022 Term] Find the sum of first 20 terms of an AP, whose nth term is given as @, =5~2n- In an AP, if S; +S, = 167 and So = 235, then find the AP, where S, denotes the sum of its first mth terms. [cose 2015] How many multiples of 4 lie between 10 and 205? INCERT EXERCISE; C8SE2019] How many integers between 200 and 500 are divisible by 8? [BSE2017}Q15. IFS, the sum of first n terms of an AP is given by 5, = 3n? — 4n, find the nth term and common difference. the 30th instalment? What amount of loan has he paid after 30th instalment? [BSE 2023} [CBSE2019} 3. Solve the equation for x:1+4+7+10+..4= @ Short Answer Type-ti Questions wy 287. INCEnT ExENPLAR; CBSE 2023, 20) Q14. Show that the sum of all terms of an AP whose first QL How many terms are there in AP whose first and fifth terms are ~14 and 2, respectively and the last term is 62. case 2023} Q2. Which term of the AP : 65, 61,57, 53,.....is the first negative term? [case 2023) 3. The 24th term of an APis twice its 10th term. Show that its 72nd term is 4 times its 15th term. 4. Ifthe 3rd and the 9th terms of an AP are 4 and -8 respectively, which term of this AP is zero? INCERT EXERCISE; V. Imp] QS. The sum of the Sth and the 9th terms of an AP is 30. fits 25th term is three times its 8th term, find the AP, [W.tmp,] Q6. Each year, a tree grows 5 cm less than it grew the preceding year. If it grew by 1 m in the first year, then in how many years will it have ceased growing? [cose 2015] 1 Q7. Find the number of terms of the AP: 18,155, 13, 1 495 and find the sum of alts terms. 8. The sum of first 15 terms of an AP is 750 and its first term is 15. Find its 20th term. [case 2023) Q9. The first term of an APis 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference. [NNCERT EXERCISE; CBSE 2017), 10. How many terms of the AP: 54,51, 48,... should be taken so that their sum is 513? Explain the double answer. QL If the sum of first 7 terms of an AP is 49 and that of its first 17 terms is 289, find the sum of first terms of the AP. [cese2019, 17,16] 12. Rohan repays his total loan of 118000 by paying every month starting with the first instalment of % 1000. If he increase the instalment by % 100 every month, what amount will be paid by him in term is g, the second term is 6 and the last term is (a+0)(b+¢~2a) 2(b-a) INCERTEXEMPLAR; CBSE 2020] «is equalto Q15. Find the sum of all 11 terms of an AP whose middle term is 30. [case 2020} @ Long Answer tupe Questions y QL. The first term of an AP of 20 terms is 2 and its last term is 59. Find its 6th term from the end. [case 2017] Q2. Find the number of terms of the AP: »21.1f Lis added to each term of this AP, then find the sum of all terms of the AP thus obtained. Q3. IFS, denotes the sum of the first n terms of an AP, prove that S55 = 3 (Spo Syo)- Q4. If the sum of first 6 terms of an AP is 36 and that of the first 16 terms is 256, find the sum of first 10 terms. [cose 2023} QS. Find the 60th term of the AP: 8, 10, 12,...if it has a total of 60 terms and hence find the sum of its last 10 terms. [cose 2015} Q6. Ifthe sum of the first p terms of an AP is the same as the sum of its first q terms (where pq), then show that the sum of first (p + q) terms is zero. [case 2019} Q7. Inan AP of 50 terms, the sum of the first 10 terms is 210 and the sum of its last 15 terms is 2565. Find the AP. [c8sE 2017; V. tmp.) Q8. The ratio of the 11th term to the 8th term of an AP is 2 : 5. Find the ratio of the Sth term to the 21st term. Also, find the ratio of the sum of first 5 terms to the sum of first 21 terms. [cose 2023} Solut ns Very Short Answer Type Questions 1. Given. AP sequence is /3.Vi2.V27.V46. or V3.23.3V3.43, Now. common difference. de Second term - First term = 23-3 23 2. Given, a=3,d=—Zand{=-11 TRICK nth term of an AP is:0,=l=a+ (9-1) d =1123+(n=1)(-2) = =Me(n-I-2) = Jan=1 = eB Hence, number af terms exist in an AP is 8. BL Let the first term and common difference of an AP be cand d respectively.Given, on)- = [0+ (21-14) [0+ (7-1)¢)=84 (nth term of AP. 0,= 0+ (n=1}d) > 0+ 20d-0-6d=B4 = 1ad= 84 = =84 5 ge a= d=6 4, Leta be the fist term and 'd' be the commen aifference of AP Given, seventh term (o,) =4 av(7-Nde4 (nth term of AP, 0,= a+ (n=I)d) = 046 (-4)=4 (ed=~4. given} = 0-24 =4 0=28 Hence. required first term Is 28 5. theven apis 15,124 Here. first term (a) = 18 and common difference (d) = 15 —18 3 jg 31-36_- = I ome Suppose there are n terms in the given AP. Then, a,=—47 rth termof AP. 9,=0+(n—1)d 47-184 (0-9/ = ~47-18 > 65 = 13 = =5n = n=2.27 5 Hence. there are 27 terms in the given AP. 6. The two-digit numbers divisible by 3 are 12.15, 18.93 Here, common difference =15-12=18-15=3 So. It forms an AP. Here, o=12,d=3and(=99 one l=a+(n=1d 99=12+(a-1)3 => 3(n-)=87 = n-1=29 > n=30 Hence. the two-digit numbers divisible by 3 are 30. TRICK pth term from the end = (n—p + 1th term from the beginning, where n is the number of terms. Given AP Is 5, 9,13, Here, a= 5,d=9-5=4and,= 185 185, a+(n-I)d 185 =5 +(n-1) (4) = 180=4(n-1) > n- 45 => nea6 Now. Sth term from the end = (46 - 9 + 1)th term from the beginning = 38th term fram the beginning $0837d=5+37%4 (: 0,=0+(n-1)d) 5+148=153 Hence, Sth term from the end Is 153, @. Given, a» 2, d=3and n= 10. TRICK The sum of first n terms of an AP is S, S20 + (0-1), Sip = Plax 2+(10-1)x3) = {449 x3) 5(4 +27) = 5x 318155 Hence, sum of first 10 terms of an AP is 15. 9. Let the sum of n terms of an AP be zero. je. 5,=0. Here, a= 1B and d= 16-1 2 5, $l20-(n—Md) ” O=F[2«18+(n—1)(-2)] [fromeg. (1)) = n(36-2n+2)=0 = 38n-2n?= But 1 = 0 (not possible) Hence, the required number of terms is 19. {EOMMON] ERROR ‘Some students consider both values of n in the answer but itis wrong approach. So, be careful about this. 10. Given, nth term of an AP Is Ty= (ana) Here, = T= 2(1)+3=5 12 1,=2(5)+3=13 TRICK The sum of n terms of an AP is S, Ste +0), where Lis the last term of an AP. P n= Blo) 5 5 $= $15 +13] = 3x18 = 45 Hence. sum of first five terms of an AP Is 45. Short Answer Type-1 Questions 1. Let nth term of the AP be - 150. Here. o= 17. d= 12-17 =—5 and a,=—150. ath term of AP. a,-a+(n-1)d =150 = 17+ (n=1) (5) = -150=17-5n+5= Sm=150+ 22-172 = W344 (isnot natural number) Hence. -150 is not a term of glven AP, {EOMMONJ ERROR Sometimes students consider as round off value of n, ie, 34 is the answer, but it is wrong. Since, 'n' is always @ natural number, so we con't round off the value of n. 2. Given. AP-8.14,20.26. Here, first term (a) =6 and comman difference (d) = 4-8 =6 Letts nth term will be 72 more than its 4st term, y= 0q 472 = a4 (n=1)d=0+ (41-1) 472 [: a,=0+ (n=) = (n-1) 6 =40 x6 +72 ne 53 So, its 53rd term is the required term, 3. Let and dbe the first term and common difference of an AP. The according to the given condition. 0,+0,=0 TR! CK- nth term of an AP +(a- Yo (a+ (3-Yo}+[0+(7-1)d)=0 = o+2d+a46d=0 = 20+8d=0 = odd =0 = 0+ (5-1) d=0 = 05-0 Hence. Sth term is zero. 4, Let aand dbe the first term and common difference of an AP, Then, Tx Tye 5x Ts Tx (0+ (7-Nd} 25x [04 (5-d) fe T= 0+ (od) > Tla+6d}-5{o+ 4d) = 7o-50=20d-42d = 2a--22d = a=-Nd Tg=0+(I2-1N)d =-ld+id=0 5. Let aand de the first term and common difference of an AP. Then, a,=5 and o=9 TRICK The nth term of an APs given by a, 0-+ (n—1)d = o+(3-Td=5 and o+(7-I)0 a+2d=5 and a+ 6d: u On solving we geta#3 and dol TiP The series of an AP is a,0 + d,a + 24,0 + 3d, = The series of an APIS 3,3+1.3+2.3+3 12,3.4,5.6, 6. Given. first AP: 63. 65. 67. Here. First term (0) = 63, and common difference (d) = 65 - 63=2 Now, ath term, a,= 0% (n=1)d 33+ (n= 1) 22+ 61 Second AP: 3,10. 1 Here, first term (a’) « 3 and common difference (¢)=10-3=7 Now, ath term.a,=a' + (n=N)d" =34(n-1)7=7=4 According to the question. a6, = 2n+61=7n-4 = 5n=65=> n=13 7. Let o be the first term and d be the common difference of given AP. Given. 3rd term of AP. ay = 16 TRICK nth term of AP.a, = 0 + (n— 1)d, where a and d are the first term and common difference respectively 0+(8=Nd=16 > a+ 2d=16 a According ta the question. 0)=12+05 = 0-95 > (o(7=Nd)=[0+(5-Na)=12 > (0+ 60)-(0+ 4d) =12 = 20. de6 Put d= 6in eq (I). we get 0+ 2(6)=16 = a+ T2=16 as 4 Hence, AP will be o: (0d), (o¥ 2d), (o# 3d), le. 4, (6 +6). (4426), (443 «6). le 4,10, 16. 22 B. Let o be the first term and d be the common aifference of the given AP Here.a=6, d=13-6=7and o,=(=216 nth term of AP. p= l= 0% (n=1)d (l) > 216 =6 + (n=1)7 > 210 = (n=17 = n=1=30 = n=31 Middle term of the given AP = Pint ) 32 1 = do+y- = 16th term From eq, (1). = 0+ (16=1)d 6+15x7=6+105-™ Hence. the required middle term is 11.9. Given AP sequence is 30, -24, 16, Let @ be the first term and d be the common difference of given AP. Here a=-30,d=-24+30=6 TRICK The sum of n terms ofan APIs: S, =Fa+(n-1)0) The sum of 30 terms of an APIs 5y9=7P(2x(-30) +(30-1)6) =15(-60+174) =15x114=1710 10. Given. 5, =n(anet)=4n?-n TRICK iith term of an AR whose sum Is S,, is %=Sq-Spaa 0, =4n? +n-[4(n—1}? +(n-1)) 4n? +1 [4(n? +1-2n)+(n-1)) 4n? +n (an? +4~Bn+n-T} 24n? +4n=(4n? -7n+3) = 8-3 The AP series Is 0}-03.03 The required AP series is 8(1)-3. 8(2)~3. 8(3)-3. le. 5.13.21, TL. Given, nth term of an AP is 9, =5-2n 5 -2(1)=3 0, =5-2(2)=1 03 =5-2(3)=—1 Here iret a3 and d=0,-0,=1-3 n S_=Gl20+(n-N) Sin =22t2x(3) +(20-11-2)) = 10(6 +19«(-2)} = 10(6-38)=-320 12. Let ‘o' be the first term and ‘d’ be the common difference of an AP. Given, 5545)9167 5 3120+5- Vd) +F120+ (7-1)d]=167 3 100+5x4d+ a+ x 6d= 334 = 240+ 62d=334 = 120+ 31d= 167 0) Also. Sio= 235 s Bi20+(10-1)a]= 235 = 5(20+8d) = 235 = 20+8d=47 2) On multiplying eq. (2) by 6 and subtracting it from eq, (I), we get (120+ 31d) - (120+ 54d) = 167 - 282 = -23d=-115 > 435 On putting the value of ‘din eq (2). we get 20+9%5=47 > 20445047 > 20247-4502 = 21 So, required AP is 0. (0+ d) ie, 1 (+5). (145 *2),. le, 1.6.11 (or20).2 WB. Let a be the first term and d be the common difference of an AP. TRICK First multiple of 4 which is greater thon 10 is 12 ond. next will be 16 and so on. Therefore, the multiples of 4 are 12, 16, 20, 24... When we divide 20S by 4, the remainder will be 1. Therefore, 205 — 1 = 204 will be the last number before 205 divisible by 4 ‘The sequence of multiples of 4 Ue between 10 and 205 is as follows: 12:16) 20)24,.., 204 Here, a=12.d. rth term of AP. @)=0%(n=Nd 204 212+ (n=1)4 = 204=12+4n= 4 = 204-B8=4n . Ann 196 = n=48 Hence. there are 49 multiples of 4 lies between 10 and 208. VW. Let obe the first term and dbe thecommon difference Clearly, 208 is the first number divisible by 8, lying between 200 and S00. When S00 is divided by 8, then the remainder obtained is 4, so the last number divisible by 8, lying between 200 and $00 is S00 ~ 4 = 496, st of integers divisible by 8. tying between 200 and 500 is 208, 216, 224, .., 496 It represents an AP whose common difference (d} is. Here. first term (a) = 208 and last term (o,) = 496 inth term of AP, 6,204 (0=1)¢496 = 208 + (n-1) (8) 2 496 - 208 =(n-1)8 = 2886 (n-1)8 288. 3 n-1=2836 = nei? ‘Therefore, the numbers divisible by 8 lying between 200 and 500 are 37, EOMMONJERROR Students must read the question very corefully. The question includes “integers between 200 ond 500" while, sometimes students take 200 as a first term of the AP because it is also divisible by 8, but it is wrong. So, students take the very first value of this AP as 208. 15, Given, §,= 377 —4n TiP 1p sum of terms (,) ofan APIs given, then nth term (G,) of the AP con be determined by a, = S,—S,.. and common difference by d = O, = Oy “onthitermis given by ,=55= S55 =3n? —4n-(B(n=1}*-4(n—1)) 3n? = 4n~[3(n? +1-2n)-4n+ 4) =3n? -4n-3n?-3+6n+4n-4 6-7 and common difference, d= 0,~0jp-1) =6n-7-(6(n-1)-7] =6n-7—6n+6+ 6 Short Answer Type-ll Questions 1 Let i be the first term and id’ be the common difference of an AP. Given that first term (a) =-16, ast term (0) = 62 and fifth term. o, =2 = 04(5-1)d=2 Let n terms are in AP, +(n=1) 4 62=-14+(n=1) (4) = 4(n-1)=76 > (n-1)=19 3n=20 So. required number of terms is 20 2. Let nth term of the AP: 65. 61, 57. 53. negative term, Here, firct term (0) = 65 and common difference (d) = 61- 65 =: 0,= 0+ (n=1)d (o+(n—nd} <0 be the first - 65 +(0-1) (4) <0 = (65-4n+4)<0 = (69-4n) <0 = (4n-69)>0 = no 1 = nou nei8 ‘So, 1Bth term is the first negative term, [cCOMMONJERR@R Sometimes students takes the value of = 17 instead of token the votue of n = 18 3. Let ‘o' be the first term and ‘d’ be the common difference of the given AP. According to the given condition. Oy, = 2% Oy = a+ (24—Nd=2(0+010-1d) {rath term of the AP, o, = 0+ (n=1)d) > 04230 = 2x(a+3d) => 0+23d = 20+ 18d > o=5¢ ” Now. 0,= 04 (72=1)d=0+71d = 5d+7Id {from eg. (0) = Or = 76d (2) and 5 SOF Md= 504 14d (From eg, (1) = ais = 19d (3) Fram eqs. (2) and (3) its clear that On=Atimes of os [ 76d = 4x 190) Hence proved. 4. Given that 05 = 4 and Leto’ be the first term and ‘d’ be the common. difference of the given AP. ith term of AP, n= a+ (n—1)d a;=0+ (2-10 = a=a+2d “) end 0520+ (9-1) = -8=0+8d {2) ‘On subtracting eq. (I) from eg. (2), we get (a+ 8d)—(o+ 2d) «8-4 = 6d=-12 Cae4 = o=8 Let nth term of this AP be zero. Oy 0+(0=1)d O=8+(n=1) 2) > O=8-2n+2 = 2n = n=5 Hence. Sth term of this AP is zero, 5. Let ‘d and ‘d' be the first term and common difference respectively of an AP. According to the given condition, 0,+=30 = [0+ (-N)d)+[0+(8—-g = 30 Centh term of AP, 0, = 0+ (n—Td) = o+4d+a+8d=30 > 2a+12d= 30 = o+6d=15 Q) According to the question, (05 =3%03 Ss o+(25—Nd=3x(0+(8—1)d)0+ 24d = 3(a0+7d) a+ 24d=3a+21d 2o=3d = o=3d (2) On solving eqs. (I) and (2). we get 3 Sd sed=15 2 u u u 2 {from eq. (2)) S0,required APIs 0.04 d.a+24.. acne 1.3.57, 6. Given that tree grows 5 cm or 0.05 m less than Bement TiP Tree cease growing means the growth of tree becomes zero ot some stoge The following sequence can be formed 1.(1-0.05). (I= 2x 0.05).....0 ie, 1.0.95, 0.30, ., 0 which is an AP. Let aand dbe the first term and camman difference respectively of the given AP. Here, a=1, d=0.95—1=~0.05 ando, ,=0+(n=Wd O=1+ (n=1)(-0.05) => 005(n-1)=1 100 = 00-29 = n=20+1=21 Hence. in 21 years, tree will have ceased growing. 7 Given APIs 18, 154.13, ~49-4 Let gand de the first term and common difference respectively of the given AP, see er ee Here, 0=18. d= 155~18=4)-18 = and = Let the number of terms be n. nth term of AP, 0, o+(n=1)d vuuus 8 8/218 +20-1(P) “Sum of first n terms of an AP. s, =$(20+(n-N] wm[6-27eg (72-135) =7(-63) 24a Hence, sum of all given terms is ~ 44. 8. Let aand dbe the first term and common difference of an AP respectively, Given, first term (0) = 15 (20+(n-1)d) 14d=100 30 =70 d=5 O20 uu +(20-1)d [ I5+19*5 =15+95=110 So, its required 20th term is 110. 9. Let oand dibe the first term and common diference respectively of an AP. Given, a=5,(=45 and , =400 +: Sum of first n terms of an AP, Sr=Flo+t) 400 =2(5+45) 800. 50 ith term of AP, I= 0 (n= 1} 45 = 5+(I6—I)d = 40 = 15d 4 = oO -8 Hence, the number of terms is 16 and the common =4 = 400=£(50) = n=16 dtference is 10. Let aand dbe the first term and common difference respectively of the given AP. Here, a= 54, d= 51-54 =-3 and 5,=513 Let required number of terms ben ‘Then, sum of first 1 terms of the AP. = $20+(n-10) 513 = 8254 +(n-1)(-3)) 513 x Ze n(108 -3n+3) 1026 = nftN1~ 3n) 1026 = 11in- 3n? 3n?—Mn+1026=0 =37n + 342=0 mP =19n= 187 +342=0 n(n—19) ~18 (n=19)=0 (n=19) (n=18)=0 n-19=Oorn—1B=0 n=190rn=18 =. SUM of 1B terms = Sum of 19 terms But 19th term is fe Op = 54+ 18 x (3) = 54-54 =0) Hence, required number of terms is 18. (civide by 3) UUUUUUuUYCOMMON] ERR@)R Some students confused in double answer and make them mistake in writing the answer. Let ‘o’ and ‘d’ be the first term and common, difference of an AP respectively, Given, 59549 = TRov-ve)=49 “sum of first n terms of an AP, (20+(n-1d) = 2a+6d=7%2 > a+3d=7 () and Sy= 289 = Bpora-1 Fl2o+a7 Da] = 20+ 16d=17x2 > o+8d=17 (2) Subtracting eq. (I) from eq. (2). we get (0+ 8d)—(0+ 3d) =17 -7 = Sd=10->d=2 Put d=2ineq (I), we get a+3x2=7 = a-1 Now 5, Sl20+(r-V0) APQRIF(O=H(A) (-0-1andd=2) =2242n-2)=2x2n= 0? $Q+2n-2)=Bx2n= 01 Hence, the required sum Is ri? 12. instalments ta be paid by Rohan is, 1000, 100, 1200... Since. the difference between each consecutive terms is 100 (constant). 50, this sequence forms an AP. Let oand de the first term and common difference of an AP. 1200 and d= 1100-1000 =100 0,=0+(n=1)d ‘39 = 1000 + (30 - 1) (100) 1000 + 2500 = 3900 So. in the 30th instalment. he will pay @ 3900. 5, Spa+(n -)d) 30, Sup =F (21000 + (30-1) 100) = 15 (2000 + 2900) =15 x 4900 = 73500 So, Rohan has paid £73500 after 30 instalments 1B. Given, 144474104 ...+x© 287 a) TiP First of all check the series 1+ 447+ not xis in AP or Let §,=144474104..4% Here, o)=1,0,24, 0,27, 4-153 05-0)#7-423 Q)- 0)=0;~a,=3 This Series forms an AP with common ditference d=3 Let nbe the number of terms in that AP nth term of AP, On=04(A=1)d x=14(0-1)3 (nth term 0, = xand first term a= 0) =1) Now. — a~9, nakt2 Sum of first ’n terms of an AP, Flo, + 0,)- 222 (1439 -@AN2) Put the value of §, in eq, (1). we get Gas 87 > x? 4 3x42=1722 > 2 43x-172000 > x2 + 43x-40x-1720 (by splitting the middle term) > x (x43) = 40 (x +43) =0 = (x +43) (x-40) <0 = 43, 40 But x cannot be negative, because at x= - 43, nis ‘egative, which is not possible. Thus, required value of xis 40. ERROR Some students make mistake by taking both values of x as answer, but students should be remember that the number of terms n can't be negative. 14. Given, first term (A) = 0. second term =b and Last term (l) = ¢ TR!CK. nth term of ARG, =(=a+ (n= 1d where, a, = | = last term of AP and n is the number Now. ts A+(n-1d = c=0+(n=1)(b=0) ( common difference d=Hence proved. 1B. Given that. the total number of terms in an AP is equal to T1and the value of the middle most term in an AP is equal to 30 Now, here m= 11 which is odd. TR!CK net Middle term of series when n is odd = (22) term. So. middle term of AP having Titers nel (Se term = 6th term = 30 Let oand d be the first term and common difference respectively of the AP. ith term of an AP is 0, =0+(n-I)d 50% (6-1 > 30<0+5d = o=30-5d Sum of first n terms of an AP is a $= 5 (20+(n-N)d) Sy =H (2 (0-50) +(i1-I)0) (+ = 30-50) u = 4 (60)-330 260) Hence. the sum of all 11 terms of an AP Is equal to 330. Long Answer Type Questions 1 Let ‘o! and ‘d’ be the first term and common difference respectively of the given AP. Given, 9=2.[=59 and n=20 “nth term of AP, [=0,=0%(n= 1) 59 = 2+ (20-)d > 59-2 = 19d = 19d=57 52. = an Sng ith term from the end = (~ (n=I)d 6th term from the end = 59-(6-1)3 = 59-15 =44 ‘Hence, the 6th term from the end is 44. 2. Given AP is =12.-9, 6... 21 Let aand d be the first term and common difference respectively of the AP. Here, a= -12, d=—9~(-12)=3 andl=21 = 0, Let n be the number of terms in the given AB ith term of AP.(= 0,= 0+ (n=1)d 21 = 2+ (n=1)@) = 21 = 1243-3 = +15 = 3n ay or ne (On adding 1 to each term in given AP, new AP so formed is 1,-8.5... 22 Here, a= =I d= -8=(Hl) =3, n=12and (= 22 Sum of nterms of the AP, S,, 5, = (-11+22) = 6x11=66 Hence, the required number of terms Is 12 and sum of all terms of new AP is 66. 3. Let the first term and common difference of the given AP be ‘a’ and'd" respectively. ‘Sum of first n terms of the AP. 5, Fl20+ (nnd) 5 = 22420 + (30-1)¢] = S,9=15(20+29d)=300+ 435d ~{) S~ 220+ (20-04) = S910 (20+ 19d) = 200+ 180d 2 and S9= B(20+ (10-1)d) = Sq=5 (20+9d) = 100+ 45d @ Now, 3 (5;5— Sig) = 3 ((20a + 190d) — (100 + 45d)) (from eqs. (2) and (3)) (100+ 1459) 100 + 435d= S35 (frome (1)) Hence proved. 4. Let aand dbe the first term and common difference respectively of the glven AP. Given, 5. = 36 and Sys = 256 TRICK Sum of fist n terms of an AP is: S,, 6 = $120+(6-1)4)-36 and Bte0 +(16—1)d) = 256 3(20+5d)=36 and 8 (20+ 15d) =256 () (2) au and Gn subtracting eq, (I) from eq, (2). we get (20+15d)~ (20+ 5d) =32-12 a ld=20 > d=2 Put d=2 in eq. (I). we get 20+5x2=12 > 2a=12-10=2 = 0-1 Now. §,=S(20+(0—N)4)[2n) =n? Sig = (10)? = 100 Hence, sum of first 10 terms is 100 5. Given APIs 8,10, 12 Let o and dbe the first term and common difference of the AP respectively Here, o= 8 and d= 10-8 = 2 Let the number of terms in the given AP be n. rth term oF AP, 0, =0+(0-1id O49 = 8+ (60-1) 2= B+ 59x 2-84 118 = 125 OF 9126 Hence, 60th term is 126. Now. sum of its last 10 terms © Seq= Seo =S02x8+(60-1)x2]-30f2x0+ (50-1) x2] “sum of first n terms of an AP. = 30(I6 +59 x 2)—25(16 +49 x 2) = 30(16 + 118) ~ 25(16 + 98) » 30 x 134-25 x TM4 = 4020 ~2850 = 1170 Hence, sum of last 10 terms of given AP is 1170 6 Let aand dbe the first term and common difference of the AP respectively According to the question. Sy TRICK Sum of first n terms of an AP. S, =F Ra+(n ~1)d) u ${20+(p-Nd]-3[20+(9-Nd] p(20+ pd-d)= q(2a+ qd) 2ap + p*d— pd=2aq + q’d- qd 2olp- a) + dlp?— ¢?) -alp-q) =0 (p-q) (20+ d(p+q)-d=0 20+(p+q-1)d=0(- p2q)-() Now. sum of first (p + g) terms &Spig 22d) poxeeqend vuuuy (from eg (1)) st Hence proved. 2 TiP Colculate 36th and 50th terms of AP as consider first term and last term respectively, because the sum of last 15 terms of APs given and there is no idea about a’and ad'of given AP Let the first term and common difference af an AP be’o’ and 'd’ respectively. Given, number of terms in this AP, n= SO Sum of frst 10 terms of this AP, 5210 = Beoa+(i0-na}=210 Sum of first n terms of an AP, [20+(n-1}d] = 20+8d=42 0 Now. 36th term of this AP, O55 = 0+ (36 I]d= 0+ 35d Ge ath term of an AP: 0,= a+ (2-1) d) and 50th term of this AP. as = 0+ 49d ‘Sum of last 15 terms of this AP = 2565 fay, +059) = 2565 [5-3-0] = Flor 35d 10+ 48d) = 2565 = 20+84de 1x2 - 0+ 42d=171 2 Put the value of ‘o'from eq. (2) in eq. (I). we get (I= 42d) + Sd-= 42 = 75d=300 = d=4 Put the value of ‘din eg, (2), we get o+16B=1 => O83 So, required AP Is: aordar2da+3d 3,3+4342%4,3+3%4 or 3.7.11.15. 8, Let oand dbe the First term and commen difference of an AP respectively. nth term of AP. 0,,= a(n ~1)d eTith term of AP, ay = 0+(I=1) d= 0+ 10d and 18th term of AP. og 0+ (18 -N)d= a+ 17d Given, 2n.2 _, 410d _2 Oe 3 avd 3 = 30430d=204 34d = o=4d 0) Sthtermof AP _o+(5—Ild _o+4d FisttermaF AP” o=(2i=Tyd ~a+300 [om eM) Sum of first n terms of AP Is s, = Fl20+(n-Va) 5 5 320 +(5-Nd)= $2 dec Zed 4d)-312d~30d —{fromea.() and Sy, =21(20 + (21-Mhd}= 224d + 20d) 2h 204) -2-284-2944 (Homes)Hence, required ratios are a, : a = 3: 1 and Sg: Sq 5:49 pen Chapter Test Multiple Choice Questions Q1. If 7th term and 13th term of an AP are 34 and 64 respectively, then its 18th term is: 2.89 bso c92 4.94 If the sum of m terms of an AP is 3n? + and its common difference is 6, then its frst term is: Pei ba cs a6 Assertion and Reason Type Questions Directions (Q. Nos. 3-4): in the following questions, a statement of Assertion (A) is followed by a statement of Reason (A). Choose the correct option: a. Both Assertion (A) and Reason (R) are true and Reason (isthe correct explanation of Assertion (A) b. Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A) Assertion (A) is true but Reason (R) Is false d. Assertion (A) is false but Reason (R) is true Assertion (A): The common difference of an AP in which a35 — dy9= 30is 6. Reason (R): The nth term of the sequence 8,13, 18, 5043. Assertion (A): The sum of the series with the nth term T,, = 4 -2n is -208, when number of terms is 16. Reason (R): The sum of AP series is determined by Sq = Fie +(n=1)d). a2. Q3. Q4. Fillin the Blanks Q5. In any arithmetic progression, if each term is increased by 3, then the new progression series is FOTMEM rnennnee If the common difference is . then each term of the AP will be same as the first term of the AP. True/False Q7. If nth term of an AP is a,, then the common difference is determined by d= a, — ay... QB. Asequence follow certain rule is a progression. Case Study Based Question Q9. There is a great demands of electrical appliances (ie,, Freeze, Television, Cooler ete.) an electrical appliance ‘manufacturing company decided Qo. to increase its production. In every five years, the company doubles its increased production. Following the same process of increasing production, the production of company in its Sth year was 10000 sets, in the 6th year it was 11000 Sets and so on. Based on the above solve the following questions (i) Find the production of the company during first year. (li)_In which year, the production is 20,000 sets? Or Find the sum of production during first 9 yr. (ii) In how many years, company produce 2,16,000 sets? Very Short Answer Type Questions Q10. Find the 6th term from the end of the AP: 17,14, 11,, 40. QU. Find the sum of 20 terms of the AP: Short Answer Type-I Questions Q1z. Which term of the arithmetic progression 5,15, 25, ill be 130 more than its 34st term? Short Answer Type-II Questions Q13. Aman repays a loan of $5250 by paying ¢20 in the first month and then increases the payment by 15 every month. How long will it take him to clear the loan? Long Answer Type Question Q14. The ratio of the sums of m and n terms of an AP is m? : n?. Show that the ratio of the mth and nth terms Is (2m 1) : (2m 1). . information, 14,7,