Solution

PRACTICE TEST

Class 10 - Mathematics
1. Let first term of an A.P. be a.
Common difference of an A.P be d.
then, t4 = a + (n - 1)d [n = 4]
t4 = a + (4 - 1)d
⇒ -15 = 9 + 3d ...(i)
tg = 9 + (n - 1)d [n = 9]
tg = 9 + (9 - 1)d
⇒ -30 = 9 + 8d ...(ii)
On Solving equation (i) and (ii)

d = -3
Put the value of d in equation (i)
a + 3d = -15
⇒ a + 3(-3) = -15

a = -15 + 9
a = -6
Sn = {2a + (n - 1)d}
n

Sum of first 17 terms

S17 = {2 × (-6) + (17 - 1) × (-3)} [n = 17, a = -6, d = -3]
17

S17 = 17

2
{-12 - 48}
17
S17 = (−60)
2

∴ Sum of first 17 ferms if -510.​​

2. Let a - 3d, a - 2d, a - d and a are four numbers in AP.
(a - 3d) + (a - 2d) + (a - d) + a = 50
⇒ 4a - 6d = 50

⇒ 2a - 3d = 25 ...(i)

According to question,
⇒ a = 4(a - 3d)

⇒ a - 4a + 12d = 0
⇒ -3a + 12d = 0
⇒ 3a = 12d

a = 4d
Put the value of a in equation (i)
2a - 3d = 25
⇒ 2 × 4d - 3d = 25

⇒ 5d = 25

∴ d = 5

a = 4d
= 4 × 5 = 20
The numbers are 5, 10, 15 and 20​​.
3. 110, 121,... 990
Numbers which are greater than 100 and multiple of 11 forms an A.P.
first term = a = 110
common difference = d = 11

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 an = a + (n - 1)d
⇒ 990 = 110 + (n - 1)11
⇒ 990 = 11n + 99

⇒ 11n = 891
∴ n = 81

Sn = [2a + an]n

⇒ Sn = 81

2
[110 + 990] = 81

2
× 1100 = 44550
Hence, the sum of all three digit natural number which are multiple of 11 is 44550.​​
4. Let the terms of a G.P is a1, a2, a3, -an.
Here,
a2 −2 a3 −1 a2 −1
i. a1
=
4
, a2
=
1

−2
=
2
, a1
=
2
1
−
a4 2 −1
= =
a3 1 2
a2 a3 a4 −1
= = =
a1 a2 a3 2

So, we can say that sequence will be G.P.

a2 −2 −1
​Common ration (r) = a1
=
4
=
2
2
3a

a2 3a
ii. a1
=
a
4
=
4
3
9a

a3 3
16 9a 4 3a
= = × =
a2 3a2 16 3a2 4

4
a2 a3 3a
= =
a1 a2 4

So, Progression will be G.P.

2
3a

a2
Common ration = a1
=
4

a
,
3a

4
.
5. i. Given AP is 3, -2, -7, -12,...
Here, a1= 3, a2 = -2, a3 = -7, a4 = -12 and so on.
∴ Common difference (d) = - a2 - a1 = -2 - 3 = -5
ii. Given AP is 11, 11, 11, 11,...
Here, a1 = 11, a2 = 11, a3 = 11, a4 = H and so on.
∴ Common difference (d) = a2 - a1 = 11 - 11 = 0
1 1 1 1
iii. Given AP is 5 2
,9
2
, 13
2
, 17
2
,...

Here, a1 = 5 1

2
, a2 = 9
1

2
, a3 = 13
1

2
, a4 = 17
1

∴ Common difference (d)

1 1 19 11 8
= a2 - a1 = 9 2
− 5
2
=
2
−
2
=
2
=4
6. Let a

r
, a, ar be a G.P.
a 13
+ a + ar =
r 12
2
a+ar+ar 13
⇒
x
=
12
...(i)
a
× a × ar = −1
r

⇒ a3 = -1
∴ a = -1
Put the value of a in equation (i)
2
(−1)+(−1)r+(−1)r 13
⇒ =
r 12
2
−1−r−r 13
⇒ =
r 12

⇒ 13r = -12 - 12r - 12r2

⇒ 12r2 + 25r + 12 = 0
12r2 + 16r + 9r + 12 = 0
4r(3r + 4) + 3(3r + 4) = 0
(4r + 3) = 0 and 3r + 4 = 0
−3 −4
r= 4
and r = 3

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 −3 −1 −3
If r = 4
, then −3
, -1 and (-1)( 4
)

3
, -1 and 3

4
are in G.P.
−4
If r = 3
, then 3

4
, -1 and 4

3
are in G.P.​​
7. Let her pocket money be ₹x.
Given, she puts ₹1 on day 1, ₹2 on day 2, ₹3 on day 3 and so on till the end of the month, into her piggy bank. Clearly, the
amounts that she puts everyday of the month, form an AP, in which number of terms is 31, first term (a) = 1 and common
difference (d) = 2 - 1 = 1
Now, sum of first 31 terms,
31
S =31 [2 × 1 + (31 - 1) × 1] [∵ Sn =
2
n

2
{2a + (n - 1) d}]
=
31

2
(2 + 30) =
31×32

2
= 31 × 16 = 496
So, Kanika puts ₹496 into her piggy bank.
Also, it is given that, she spent ₹204 of her pocket money and found that at the end of the month, she still had ₹100 with her.
According to the condition,
(x - 496) - 204 = 100
⇒ x - 700 = 100
⇒ x = ₹800

Hence, ₹800 was her pocket money for the month.

8. Given, a4 = 0
a + (n - 1)d = 0 [n = 4]
a + 3d = 0
a = -3d
From LHS, a25 = a + (n - 1)d
= a + (25 - 1)d
= a + 24d [∵ a = -3d]
= -3d + 24d = 21d
From RHS, a11 = a + (n - 1)d
= a + (11 - 1)d = a + 10d [∵ a = -3d]​
= -3d + 10d
= 7d
∴ Hence, a25 = 3 × a11

9. a = 1 = First term of an A.P.

last term = an = 20
Sum of n terms of an A.P, S n =
n

2
[a + an ]

n
Sn = 2
[first term + last term]
⇒ 399 = n

2
[1 + 20]
399 = n

2
× 21

399×2
∴ n=
21
= 38
The value of n is 38.​​
10. Let the G.P be a, ar, ar2, ar3, ... arn-1
First term = a
nth term arn-1
But nth term = b (Given)
∴ b = arn-1
Now, product of n terms of GP is given which is equal to p.
∴ P = (a)(ar) (ar2) (ar3) ... (arn-1)
P = an(1 × r × r2 × r3 × ... rn-1)
P = an(ro × r1 × r2 × r3 × ... rn-1)
P=an r(0+1+2+3+...+(n-1)+n)
(n−1)(n)

n
p = a r 2

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 Now; squaring both sides
−2(n−1)(n)

P2 = a 2n
r 2

P2 = (a2 rn-1)n
P2 = (a⋅ arn-1)n
P2 = (a⋅ b)n {∵ arn-1 = b}
​Hence proved.
11. Multiple of 7 which are greater than 500 = 504
Multiple of 7 which are less than 900 = 896.
First term = a = 504
Common difference = d = 7
Last term = an = 896
an = a + (n - 1)d
⇒ 896 = 509 + (n - 1)7
⇒ 896 = 500 + 7n - 7
⇒ 7n = 399
∴ n= = 57 399

no. of term = n = 57
S = n [29 + an]
n

= n

2
[First term + last term]
= 15

2
[504 + 896]
⇒
57

2
× 1400 = 39900
Hence, the sum of all multiple of 7 lying between 500 and 900 is 39900.​​
12. ∵ We have to find sum of later half of the 2n terms.
It means we have to find last 'n' terms of 2n terms of an AP.
i.e. {nth + (n + 1)th + (n + 2)th... (2n)th} term
Above sum can also be found out as adding the sum of first 'n' terms and subtracting the same, we get
= {1st + 2nd + 3rd + 4th + ...(2n)th} + {1st + 2nd + 3rd +...+ nth}term
= (2a + (2n - 1)d) - (2a + (n - 1)d)
2n

2
n

2
n
Taking 2
common
= n

2
[4a + 2(2n - 1)d - 2a - (n - 1)d]
= n

2
[2a + 2(2n - 1)d - (n - 1)d]
n
= 2
[2a + d(4n - 2 - n + 1)]
= n

2
[2a + (3n - 1)d]
On multiplying and dividing by 3, we get
=
1

3
[
3n

2
(2a + (3n − 1)d] ...(i)
Hence, the expression written in equation (i) represents one third the sum of first 3n terms.
Hence, it is proved that sum of later half of 2n terms any A.P is rd the sum of 3n terms of the same AP.
1

13. i. First term (a) = 3 (Given)

Last term (Tn) = 96 (Also given)

∴ Tn = arn - 1 (By using formula)

⇒ 96 = 3(2)n - 1
Here, common ratio (r) = 2 (given)
(2)n - 1 =
96
∴
3
= 32
i.e., (2)n - 1 = (2)5
Comparing the powers, we get n - 1 = 5
⇒ n = 6

Hence, the number of terms of the G.P. is 6.

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 n
a( r −1)
ii. ∴ Sum of n terms of the G.P. (Sn) = r−1
6
3(2 −1)

i.e., S6 = 2−1
[Taking values a, r and n from part (i)]
3(64−1)
= 1

= 3 × 63 = 189

14. Tn = a(rn - 1)

T10 = a(r10 - 1) = ar9

T13 = a(r13 - 1) = ar12

T10 = 8 × T13

ar9 = 8(ar12)
r3 =
3
1 1
= ( )
8 2

∴ r= 1

2
a
S∞ =
1−r

Given, a = 3
3 3 3×2
∴ S∞ = = = = 6
1 1 1
1−
2 2

S∞ =6
5 5
15. i. Sequence: -5, − , 0, 2 2

d=− 5

2
− (−5)

d=− 5

2
+ 5

5
d= 2

tn = a + (n - 1)d
−5
t10 = 2
+ (10 − 1)
5

5 45
= − +
2 2
−5+45
=
2

t10 = 20
Hence, 10th term = 20
ii. Sequence: -3, − , 2, ... 1

2
−1
d= 2
− (−3)

1
= − + 3
2

d= 5

tn = a + (n - 1)d
t11 = -3 + (11 - 1) 5

5
= −3 + 10 ×
2

t11 = 22
Hence, 11th term of AP = 22
iii. A.P = 7, 5, 3, 1...
a = 7, a + d = 5, a + 2d = 3 are terms of A.P
d = 5 - 7 = -2
tn = a + (n - 1)d
t23 = 7 + (23 - 1) - 2
= 7 + (-44)
t23 = -37

hence, 23rd term of A.P = -37​​

a
16. Let the length, breadth and height of rectangular solid block be r
, a and ar, respectively.
= 216 cm3 [∵ Volume of a cuboid = length × breadth × height]
a
∴ Volume = r
× a × ar

3 3 3
⇒ a = 216 ⇒ a = 6

∴ a = 6 cm

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 2

Now, surface area = 2 ( a

r
2
+ a r + a )
2
= 252 [∵ surface area of cuboid = 2 (lb + bh + hl)]
⇒ 2a
2
(
1

r
+ r + 1) = 252
2

⇒ 2 × 36 (
1+ r +r

r
) = 252

1 + r2 + r =
2
1+ r +r
⇒
r
= 252

2×36
⇒
126

36
r

⇒ 1 + r
2
+ r =
21

6
r ⇒ 6 + 6r2 + 6r = 21r
2 2
⇒ 6r − 15r + 6 = 0 ⇒ 2r − 5r + 2 = 0
1
⇒ (2r − 1)(r − 2) = 0 ⇒ r =
2
or 2
For r = 1

2
, length = a

r
=
6×2

1
= 12 cm breadth = a = 6 cm
and height = ar = 6 × 1

2
= 3 cm
a 6
For r = 2, length = r
=
2
= 3 , breadth = a = 6 cm
and height = ar = 6 × 2 = 12 cm
17. Let first term of a G.P and common ratio is a and respectively.
tn = a - rn-1

ts = a - r5-1
80 = ar4 ...(i)
t8 = a⋅ r8-1

640 = a⋅ r7 ...(ii)
(i) divided by (ii)
4
a⋅r 80
=
7 640
a⋅r
1 1
⇒ =
3 8
r

⇒ r3 = 8
3 –
∴ r = √8

r=2
Put the value of r in equation (i)
a ⋅ r = 80
4

⇒ a ⋅ 2 = 80
4

80
∴ a= = 5
16

a=5
a, ar2, ar2, ar3,... is a geometric series.
5, 5 × 2, 5 × 22, 5 × 23,...
Hence, 5, 10, 20, 40,... is a geometric series.​​
18. 32, 36, 40,.... 2000
36 - 32 = 90 - 36 = 4
So, This is an A.P.
a = 32
d=4
According to question,
32 + 36 + 40 +... = 2000
Sn =
n

2
[2a + (n − 1)d] [∵ n = No. of months]
⇒ 2000 =
n

2
[64 + (n - 1) 4]
⇒ 2000 =
n

2
[60 + 4n]
⇒ 4n2 + 60n - 4000 = 0
⇒ n2 + 15n - 1000 = 0
n2 + 40n - 25n - 1000 = 0
n(n + 40) - 25(n + 40) = 0
(n - 25)(n + 40) = 0
n = 25, n = -40 [Not possible]
There are 25 months.​​

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19. Given: an = 4, d = -4, n = 7
Using the formula
an = a + (n - 1)d
4 = a + (7 - 1)(-4)
4 = a + 6(-4)
4 = a - 24
a = 28​​
20. Let a, a + d, a + 2d,... be an A.P.
a1 + a5 = 26
⇒ a + (a + 4d) = 26
⇒ 2a + 4d = 26

a + 2d = 13 ...(i), a + d = 13 - d
Again,
a2 × a4 = 160
⇒ (a + d) (a + 3d) = 160
⇒ (13 - d) (13 + d) = 160 [∵ (a + b)(a - b) = a2 - b2]
⇒ 169 - d2 = 160
d2 = 9
d = ±3
Put the value of d in equation (i)
​d = 3
a + 2d = 13
a + 2 × 3 = 13
a=7
if d = 3
a + 2d = 13
a + 2 × (-3) = 13
a = 19
7, 10, 13, 16, ... are in A.P
19, 16, 13, ... are in A.P
n
S7 = 2
[2a + (n - 1)d], a = 7, d = 3
7
= (2 × 7 + 6 × 3)
2
7 7
= 2
× (14 + 18) = 2
× 32 = 112
Again, S7 = n

2
[2a + (n - 1)d], where a = 19, d = -3
7
= [2 × 19 + 6 × (−3)]
2

=
7

2
× (38 − 18) = 7

2
× 20 = 70
21. Let the GP be a, ar, ar2, ar3, ...
According to the given condition,
Sum of first three terms = a + ar + ar2 = 16 ...(i)
and sum of next three terms
= ar3 + ar4 + ar5 = 128 ...(ii)
On dividing Eq. (i) by Eq. (ii), we get
2
a+ar+ar 16
=
ar3 +ar4 +ar5 128
2
a(1+r+ r )
1
⇒
3 2
= 8
ar (1+r+ r )

3 3
⇒ (
1

r
) =( 1

2
)

On comparing the base of power 3 from both sides, we get

1

r
= ⇒r=2 1

On putting r = 2 in Eq. (i), we get

a + 2a + 4a = 16

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 ⇒ 7a = 16​​⇒ a = 16

7
n
a( r −1)
Now, sum to n terms, Sn = r−1
[∵ r = 2 > 1]
16 n
(2 −1)
16
= =
7 n
(2 − 1)
2−1 7

Hence, a = 16

7
, r = 2 and S n =
16

7
(2
n
− 1)

22. Here, a = 500 gm, r = 1

T6 = ar5
5
= 500( 1

2
)

= 500

32

= 15.625 gm.
23. Given, total amount of loan = ₹3250
Amount paid in first month = ₹20
and amount increases every month = ₹15
Clearly, the amounts of repayment form an AP with first term (a) = 20 and common difference (d) = 15.
Let the loan be cleared in n months.
Then, S = 3250 n

n n
⇒
2
[2a + (n − 1)d] = ₹3250 [∵ S n =
2
{2a + (n − 1)d}]

⇒
n

2
[2(20) + (n - 1) 15] = 3250
⇒ n(40 + 15n - 15) = 3250 × 2
⇒ 25n + 15n2 = 6500
⇒ 3n2 + 5n - 1300 = 0 [dividing both sides by 5]
⇒ 3n2 + 65n - 60n - 1300 = 0
⇒ n(3n + 65) - 20 (3n + 65) = 0

⇒ (n - 20)(3n + 65) = 0
−65
⇒ n = 20 or n = 3
−65
Since, n should be a positive integer, so neglect n = 3

Thus, n = 20
Hence, the loan will clear in 20 months.
24. Let a, ar, ar2, ...be a G.P.
According to question,
a4 = (a2)2 [∵ where a = -3]
2
4−1 2−1
⇒ a⋅ r = (q ⋅ r )

3 2 2
⇒ a⋅ r = a ⋅ r

∴ a=r
a = r = -3 where r = common ratio
a7 = a⋅ rn-1

= (-3)⋅ (-3)7-1 = (-3)⋅ (-3)6 = -2187

Hence, the 7th term of a G.P. is -2187.​​
– –
25. i. √3, 3, 3√3, ... and 729
3 3√3 –
= = √3
√3 3

a2 a3 –
= = √3
a1 a2

Hence, a1, a2, a3,... an is a geometric series.

an = a1rn-1
– –n−1
⇒ 729 = √3 ⋅ √3 [∵ a
m
× a
n
= a
m+n
]

–(1+n−1)
⇒ 729 = √3
–n
√3 = 729
[∵ if am = an then m = n]
–h – 12
√3 = (√3)

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 n = 12
There are 12 terms
ii. , , , ... is
1

3
1

9
1

27
1

19683
1

9 1 3 1
= × =
1 9 1 3
3
1

27 1 9 1
= × =
1 27 1 3

9
a2 a3 1
= =
a1 a2 3

then, a1, a2, a3,... an be a G.P.

an = a1⋅ rn-1
n−1
1 1 1
⇒ = ⋅ ( )
19683 3 3

1+n−1
1 1
⇒ = ( )
19683 3

n 9
1 1
⇒ ( ) = ( )
3 3

∵ n=9
There are 9 terms.
iii. 0.2, 0.4, 0.8,... is 204.8
0.4 0.8
= = 2
0.2 0.4

a1, a2, a3,... an be a G.P

r = common ratio = 2
an = a1⋅ rn-1

204.8 = 0.2(2)n-1
⇒ 2
n−1
=
204.8

0.2
= 1024
⇒ 2n-1 = 210
n - 1 = 10
∴ n = 11
There are 11 terms.

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10 / 27
26. Let the GP be a, ar, ar2, ar3, ...
Given, third term = first term + 9

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 ⇒ ar2 = a + 9
⇒ ar2 - a = 9 ...(i)
Also, it is given that, second term = fourth term + 18
⇒ ar = ar3 + 18
⇒ ar - ar3 = 18 ...(ii)
On dividing Eq. (i) by Eq. (ii), we get
2
ar −a 9
=
3 18
ar−ar
2
a(r −1)
1
⇒ =
2 2
ar(1− r )

2
(1− r )
−1 1
⇒ =
r (1− r2 ) 2

⇒ −
1

r
=
1

2
⇒ r=-2
On putting r = - 2 in Eq. (i), we get
a(- 2)2 - a = 9
⇒ - 4a - a = 9

⇒ 3a = 9

⇒ a = 3

∴ GP is 3, 3 (- 2), 3 (-2)2, 3(- 2)3, ...

Hence, the required four numbers are 3, - 6, 12 and - 24.
27. Given,
Principal = P = ₹5000
Rate = r = 8%
p×r×t 5000×8×1
S.I. for 1st year = 100
=
100
= ₹400
5000×8×2
For 2nd year = 100
= ₹800
3rd year = 5000×8×3

100
= ₹1200
400, 800, 1200
d = common difference = 800 - 400 = 1200 - 800
= 400
Hence, 400, 800, 1200, ... are in AP
-a where a1 = 400
d = 400
an = a + (n - 1)d
a30 = 400 + (30 - 1) × 400
= 400 + 29 × 400 = 12000
Hence, the interest at the end of 30 year is 12000.
28. Given, A.P = 7, 11, 15, 19, ....
First term, a = 7
Common difference, d = 11 - 7 = 4
For nth term
Tn = a + (n - 1) d
T16 = 7 + (16 - 1)4 [∴ n = 16]
= 7 + 15 × 4
= 67
and Sn = n

2
[2a + (n - 1)d]
For n = 6
S6 = [2 × 7 + (6 - 1)4]
6

= 3[14 + 20]
= 3 × 34 = 102
Hence, 16th term of A.P = 67 and Sum of 6 terms = 102.

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29. Given, first term = 5
Common ratio = 2
Sn = 315
n
a⋅( r −1)
⇒ = 315
r−1
n
5( 2 −1)
⇒
2−1
= 315
⇒ 2n - 4 = 63
⇒ 2n = 64
⇒ 2n = 26
⇒ n = 6

No. of terms = 6
an = last term = a⋅ rn-1

= 5.26-1 = 5.25 = 5 × 32 = 160

Hence the last term is 160.​​
30. Geometic series is 5 + 25 + 125 +...
a=5
r=5
an = a⋅ rn-1

a10 = 5⋅ 510-1 = 5.59 = 510 = 97,65,625

Again, an = a⋅ rn-1

an = 5⋅ 5n-1

= 51+n-1 = 5n
Hence, the nth term is 5n.​​
31. a, ar, ar2, ... are in G.P.
Where, a = T1 = 3

an = Tn = a⋅ rn-1

⇒ 96 = a⋅ rn-1
⇒ 96 = 3⋅ rn-1
rn-1 = 32 ...(i) [ r
n
r
∴
n−1
=
r
] ​
n

[rn = 32r]
a( r −1)
Sn =
r−1

3(32r−1)
189 = r−1

⇒ 32r - 1 = 63r - 63
⇒ 31r = 62
∴ r = 2

Put the value of r in equation (i)

rn-1 = 32
⇒ 2n-1 = 25 [∵ if am = an then m = n]
n-1=5
∴ n = 6

Hence, the value of n is 6.

32. Let a - d, a, a + d are three angles in A.P.
(a - d) + a + (a + d) = 180 [sum of angles of △ = 180o]
⇒ 3a = 180
a = 60
a + d = 3(a - d)
⇒ a + d = 3a - 3d

⇒ 2a = 4d

13 / 27
 a = 2d
d= = a

2
60

2
= 30

∴ d = 30
60 - 30, 60 and (60 + 30)
30, 60 and 90 are three angles of triangle.​​
33. First multiple of 9 which is greater than 10 = 18
Last multiple of 9 which less than 300 = 297
First term = 18
Last term = 297
Common difference = 9
an = a + (n - 1)d
⇒ 297 = 18 + (n - 1)9
⇒ 9n = 288

∴ n = 32

Hence, the no. of multiple of 9 lie between 10 and 300 is 32.​​

34. Let a, a + d, a + 2d, ... are be an A.P.
an = a + (n - 1)d
a4 = 11
⇒ a + 3d = 11 ...(i)
a5 + a7 = 24
⇒ (a + 4d) + (a + 6d) = 24
⇒ 2a + 10d = 24
⇒ a + 5d = 12 ...(ii)
On Solving equation (i) and (ii) we get

1
∴ d =
2

​Hence, the common difference is 1

35. Let a1, a2, a3, ... an be a G.P.

a1 + a2 = -4

a1 + a1r = -4 ...(i) [∵ an = a⋅ rn-1]

a5 = a - a3

a1r4 = a× a1r2
4
r
= 4
2
r

r2 =4
∴ r = ±2

Put the value of r in equation (i), we get

a1 + a1r = -4
a1(1 + r) = -4
⇒ a1(1 + 2) = -4 [∵ if r = 2]
a1(3) = -4
4
a1 = −
3

Again, a1(1 + r) = -4
a1(1 + (-2)) = -4 [∵ if r = -2]
-a1 = -4
a1 = 4

if r = 2, a1 = − 4

3
then G.P a1, a1r, a1r2, a, r3,..., a1rn-1

14 / 27
 −4 4 4

3
,−
3
(2), −
3
(2)
2
,...
−4 −8 −16

3
,
3
,
3
is a G.P.
if r = -2, a1 = 4

then, a1, a1r, a1r2, a1r3,...

4, 4(-2), 4(-2)2, 4(-2)3​, ...

4, -8, 16, -32,... is a G.P
36. Since, expression for general term is given. So by putting n = 1, 2 and 3 in general term expression. We get the 1st, 2nd and 3rd
term of the sequence.
i. an = 5n + 3
a1 = 5 × 1 + 3 = 8
a2 = 5 × 2 + 3 = 13
a3 = 5 × 3 + 3 = 18

ii. an = 3n2 - 2

a1 = 3(1)2 - 2 = 1

a2 = 3(2)2 - 2 = 10

a3 = 3(3)2 - 2 = 25
n−5
iii. a n =
2
1−5
a1 = = −2
2
2−5 −3
a2 = =
2 2
3−5
a3 =
2
= -1​​
37. i. a = -2, d = -2
∵ a1 = a = -2
a2 = a + d = -2 + (-2) = -4
a3 = a + 2d = -2d + 2(-2) = -6
a4 = a + 3d = -2 + 3(-2) = -8
Hence, -2, -4, -6 and -8 are the ist four terns of the A.P.
1
ii. a = − , d = -1 2

∵ a1 = a = − 1

1 3
a2 = a + d = − 2
+ (−1) = −
2

5
a3 = a + 2d = − 1

2
+ 2(−1) = −
2

a4 = a + 3d = − 1

2
+ 3(−1) = −
7

1 3 −5 −7
Hence, the four terms of A.P − 2
,−
2
,
2
and 2

iii. a = 0.5, d = 1.2

a1 = a = 0.5
a2 = a + d = 0.5 + 1.2 = 1.7
a3 = a + 2d = 0.5 + 2(1.2) = 0.5 + 2.2 = 2.9
a4 = a + 3d = 0.5 + 3(1.2) = 0.5 + 3.6 = 4.1
Hence, the first four terms of the given first term and common difference is 0.5, 1.7, 2.9 and 4.1.​​
38. 30, 28, 26 ... stacks are arranged in a row.
Let the required no. of rows be n.
30 + 28 + 26 ... + n term = 228
Clearly, this forms an A.P where
a = 30, d = 28 - 30 = -2, Sn = 228
∵ Sn = n

2
{2a + (n - 1)d}
228 = n

2
{2(30) + (n - 1)(-2)}

15 / 27
 2 × 228 = 60n - 2n2 + 2n
2n2 - 62n + 456 = 0
n2 - 31n + 228 = 0
n2 - 19n - 12n + 228 = 0
n(n - 19) - 12(n - 19) = 0
(n - 12) (n - 19) = 0
n = 12, 19
Now, to find the no of logs in the top row using.
tn = a + (n - 1)d
for n = 19
tn = 30 + (19 - 1)(-2)
= 30 + 18(-2)
= 30 - 36
tn = -6
for n = 12
tn = 30 + (12 - 1)(-2)
= 30 + 11(-2)
tn = 8
Meaningless since no. of logs can not be -ve.
Hence, no. of rows = 12
No. of logs in top row = 8​​
39. Let the three terms of AP be
(x - d), x(x + d)
(x - d) + x + (x + d) = 33
⇒ 3x = 33
x = 11
According to questions
(x - y) (x + 0) - x = 29
x2 - d2 - x = 29
putting x = 11
⇒ (11)2 - d2 - (11) = 29
⇒ 121 - d2 - 11 = 29
⇒ 110 - d2 = 29
⇒ 110 - 29 = d2 ​
⇒ 81 = d2
⇒ d=9
Since, we have x = 11 and d = 9
AP = (x - d), x, (x + d)
= (11 - 9), 11, (11 + 9)
AP = 2, 11, 20, 29, 38...​​
40. Given, initial money P = ₹2000
Rate of interest, R = 1% per year; Time, T = 1, 2, 3, 4, ...
We know that simple interest is given by the formula
SI = P RT

100

∴ SI at the end of 1st year = 2000×7×1

100
= ₹140
SI at the end of 2nd year = 2000×7×2

100
= ₹280
SI at the end of 3rd year = 2000×7×3

100
= ₹420
Thus, required list of numbers is 140, 280, 420, ....
Here, 280 - 140 = 420 - 280 = 140
So, above list of numbers forms an AP, whose first term (a) = 140 and common difference (d) = 140.

16 / 27
 Now, SI at the end of 20 yr will be equal to 20th term of the above AP.
∵ a20 = a +(20 - 1)d = 140 + 19 × 140 [∵ an = a + (n - 1)d]

= 140 + 2660 = 2800

Hence, the interest at the end of 20 yr will be ₹2800.
n
a( x −1)
41. Sum of n terms of a G.P, S n =
−1
r
3
a(r −1)

S3 =
r−1
...(i)
6
a( r −1)
S6 =
r−1
...(ii)
According to question, we get
a( r3 −1)

s3 125 r−1
= =
s6 152 6
a( r −1)

r−1
3
a(r −1) (r−1)
125
⇒ = ×
152 r−1 6
a(r −1)

[∵ a2 - b2 = (a + b)(a - b)]
125 r −1
⇒ =
152 6
r −1
3
r −1 125
=
6 152
r −1
3
r −1 125
⇒ =
2 2 152
( r3 ) − 1

3
r −1 125
⇒ =
3 3 152
(r +1)(x −1)

1 125
⇒ =
3 152
r +1

125r3 + 125 = 152

3 152−125
r =
125
27
=
125

∴ r= 3

Common ratio is 3

5
.​​
42. a1 = first term = x + 9

a2 = 2nd term = x - 6

a3 = 3rd term = 4
a2 a3

a1
=
a2
[∵ This is G.P.]
x−6 4
⇒ =
x+9 x−6

⇒ (x - 6)(x - 6) = 4(x + 9)
⇒ x2 - 6x - 6x + 36 = 4x + 36
⇒ x2 - 12x + 36 = 4x + 36
⇒ x2 - 12x + 36 - 4x - 36 = 0
⇒ x2 - 16x = 0
x(x - 16) = 0
x = 0, x - 16 = 0
x = 16
x = 0, 16
Value of x are 0 and 16.​​
43. Let the three sides of a triangle a

r
, a and ar are in G.P.
a

r
+ a + ar = 37
2
a+ar+ar
⇒
r
= 37 ...(i)
a
= 9
r

a = 9r
Put the value of a in equation (i)
9r + 9r⋅ r + 9r⋅ r2 = 37r
⇒ 9⋅ r3 + 9r2 = 28r
⇒ 9r2(r + 1) = 28r

17 / 27
 ⇒ 9r2 + 9r - 28 = 0
⇒ 9r2 + 21r - 12r - 28 = 0
⇒ 3r(3r + 7) - 4(3r + 7) = 0

⇒ (3r - 4) (3r + 7) = 0
−7
r = , 4

3
[Not define]
3

a = 9r = 9 × 4

3
= 12 = second side
2

third side = ar = 9r2 = 9 × ( 4

3
) = 16

44. Given, 2n - 1, 3n + 2 and 6n - 1 are in A.P.

d = common difference = 3n + 2 - (2n - 1)
= 3n + 2 - 2n + 1 = n + 3
n + 3 = 6n - 1 - (3n + 2) = Common difference
⇒ n + 3 = 3n - 3

⇒ 2n = 6
∴ n = 3

The value of n is 3.
5, 11 and 17 are in A.P.​​
45. Given, total amount Sandeep had = ₹12000
In the beginning of every month, he spend = ₹500
So, in the beginning of 1st month, he had amount, t1 = ₹12000
In the beginning of 2nd month, he had amount, t2 = 12000 - 500 = ₹11500
In the beginning of 3rd month, he had amount, t3 = 11500 - 500 = ₹11000
the beginning of 4th month, he had amount, t4 = 11000 - 500 = ₹10500 and so on.
Now, the list of amounts is 12000, 11500, 11000, 10500, ...
Here, t − t = t − t = t − t = -500
2 1 3 2 4 3

i.e. the difference of any term from its preceding term is constant.
So, the above list of numbers forms an AP.
46. Let a - d, a, and a + d are in an A.p.
(a - d) + a + (a + d) = 210
⇒ 39 = 210
∴ a = 70
a(a + d) = 5110
⇒ 70(70 + d) = 5110

70 + d = 73
∴ d=3
Hence, 67, 70 and 73 are in an A.P.​​
47. Let a be the first term and r be the common ratio of the given GP. Then,
n 3n 2n
a(1− r ) a(1− r ) a(1− r )
S1 (S3 − S2 ) = ⋅ { − } [let r < 1]
(1−r) (1−r) (1−r)

n 3n 2n
a(1− r ) (a−ar −a+ar )

= (1−r)
⋅
(1−r)

n 2n n 2 2n n 2
a(1− r ) ar (1− r ) a r (1− r )
= (1−r)
⋅
(1−r)
= 2
...(i)
(1−r)

2
2n n
a(1− r ) a(1− r )
2
and (S 2 − S1 ) ={ (1−r)
−
(1−r)
}

2n n 2
(a−ar −a+ar )
= 2
(1−r)

n n 2 2 2n n 2
{ar (1− r )} a r (1− r )
= 2
= 2
...(ii)
(1−r) (1−r)

Now, from Eqs. (i) and (ii), we get

S (S − S ) = (S − S )
2
1 3 2 2 1

48. Given,
A.P = 25, 50, 75, 100...
pth term (tp) value = 1000, find p

18 / 27
 d = 50 - 25 = 25
tp = a + (p - 1)d
1000 = 25 + (P - 1)25
1000 = 25 + 25p - 25
1000

25
=p
p = 40
i.e. hence, value of p = 40​​
49. Let the terms be a - d, a, a + d
∴ a - d + a + a - d = 42

3a = 42
∴ a = 14
(a - d)(a + d) = 52
142 - d2 = 52
d2 = 196 - 52
d2 = 144
∴ d = ± 12
d = 12, or -12 N.P.
First term is 14 - 12 = 2 or 14 + 12 = 26.
50. 1st generation = 2
2nd generation = 4
3rd generation = 8...
10th generation = 9
Where, a1 = 2 = first term
r = common ratio = 2
n
r −1
Sn = a (
r−1
) [∵ where r > 1]

= 211 - 21 = 2046
10
2 −1
S10 = 2 ( )
2−1

Hence, There are 2046 ancestors during the ten generations preceding his own.​​
51. 1 + 3 + 32 +... = 3280
2
3 3
= = 3
1 3

So, This is a G.P

Where, a = 1, r = 3
an = a⋅ rn-1

= 1.3n-1 = 3n-1
n
a( r −1)
Sn =
r−1
n
1( 3 −1)
3280 = 3−1

⇒ 3n - 1 = 6560
⇒ 3n = 6561
⇒ 3n = 38 [∵ am = an then m = n]
∴ n = 8

Hence, there are 8 terms.​​

52. The number which is divided by 4 and leave a remainder 3 = 11
a1 = first term = 11
Again, an = last term = 299
The numbers 11, 15, 19,... 299 are in an A.P.
Where, a1 = first term = 11
d = common difference = 4
an = a1 + (n - 1)d
⇒ 299 = 11 + (n - 1)d

19 / 27
 ⇒ 299 = 11 + 4n - 4
4n = 292
∴ n= = 73 292

Hence, there are 73 numbers lie between 10 and 300, which divided by 4 and leave a remainder 3.​​
53. We have, Sn = 6n + 7n2 ...(i)
On replacing n by (n - 1)in Eq. (i), we get
= 6(n − 1) + 7(n − 1) = 6n − 6 + 7 (n
2 2
Sn−1 + 1 − 2n)

[∵ (a - b2 = a2 + b2 - 2ab]
= 6n - 6 + 7n2 + 7 - 14n = 7n2 - 8n + 1 ...(ii)
Clearly, nth term, a = S − S n n n−1

= 6n + 7n2 - 7n2 +
8n - 1 = 14n - 1
Now, 10th term of the AP, [from Eqs. (i) and (ii)]
a10 = 14 × 10 - 1 = 140 - 1 = 139
54. Let a, a + d, a + 2d, .... are in A.P.
a = first term = 5
Sn = [2a + (n - 1)d]
n

S4 = 4

2
[2 × 5 + 3d]
= 2(10 + 3d) = 20 + 6d
8
S8 = [2 × 5 + 7d]
2

= 4[10 + 7d] = 40 + 28d

According to question,
S4 = [S8 - S4]
1

1
⇒ 20 + 6d = 2
[40 + 28d - (20 + 6d)]
⇒ 20 + 6d = 10 + 11d
⇒ 5d = 10
∴ d = 10
Common difference = d = 10​​
55. Sum of n terms of a G.P.
n
a(1− r )
Sn =
1−r
, where a = first term
r = common ratio
Sum of (n + 1)th to (2n)th term
Sn = S2n - Sn
2n n
a(1− r ) a(1− r )
= −
1−r 1−r

=
a

1−r
[1 - r2n - 1 + rn]
n
a n n ar n
= [r (− r + 1)] = (1 − r )
1−r 1−r
n
a(1−r )

Sn
Ratio = =
a⋅r
n
1−r

(1−r
n
)
Sn

(1−r)
n
a(1− r ) 1−r 1
× = n
n n r
(1−r) a⋅ r (1− r )
sn
Hence, sn
=
r
1
n
​
56. 12, 18, 24, ..., 96
Where first term = 12
Last term = 96
an = a + (n - 1)d [∵ d = common difference = 6]
⇒ 96 = 12 + (n - 1)6
96 = 12 + 6n - 6
6n = 90
∴ n = 15
Hence, there are 15 numbers which are divisible by 6.​​

20 / 27
57. Given, a1 = 10
d=5
nth term = an = a + (n - 1)d
a100 = 10 + (100 - 1)5
⇒ a100 = 10 + 99 × 5
= 505
Here, 10, 15, 20, ... 505 is an A.P.
from end,
First term = 505
Common difference = -5
a50 = a + (50 - 1) - 5
= 505 + 49 × (-5)
= 505 - 245 = 260
Hence, 50th term from the end is 260.​​
58. Let the first term of given G.P. be a and common ratio be r.
given 1st term a = 4
and 8th term T8 = 512

Tn = arn - 1

∴ T8 = ar8 - 1 = 512

4r7 = 512
r7 =
512

4
= 128
−−−
r=
7
√128

r=2
So the common ratio = 2
n
a( r −1)
Sn = r−1

5
4(2 −1)

S5 = 2−1

4(32−1)
= 1

= 4 × 31 = 124
i. Hence, common ratio = 2
ii. Sum of 5 terms = 124
59. Given, a1 + a5 + a10 + a15 + a20 + a24 = 220
We know that, an = 9 + (n - 1)d
So, 95 = 9 + (5 - 1)d = 9 + 4d
a1 + a5 + a10 + 915 + 920 + 922 = 225
a + (a + 4d) + (a + 9d) + (a +14d) + (a + 19d) + (a + 23d) = 225
⇒ 6a + (4d + 9d + 14d + 19d + 23d) = 225

⇒ 6a + d(4 + 9 + 14 + 19 + 23) = 225

⇒ 3(2a + 23d) = 225

⇒ 2a + 23d = 75

S = n {2a + (n - 1)d}
n

S24 =
24

2
{2a + (24 - 1)d}
⇒ 12(2a + 23d) [∵ 2a + 23d = 75]
= 12 × 75 = 900
Hence, the sum of first 24 terms of the AP is 900.​​
60. Let the three terms of G.P. be , a and ar a

r
39
Since, sum of three terms of G.P is 10
39
a

r
+ a + ar =
10
...(i)
a
(
r
) (a) (ar) = 1 ...(ii)

21 / 27
 a3 = 1
a = 1 taking real roots only
Now, putting value of a = 1 in equation (i)
1 39
⇒ + 1 + r =
r 10
2
1+r+r 39
⇒ =
r 10

⇒ 10 + 10r + 10r2 = 39r

⇒ 10r2 - 29r + 10 = 0
⇒ 10r2 - 25r - 4r + 10 = 0
⇒ 5r(2r - 5) - 2(2r - 5) = 0

⇒ (5r - 2) (2r - 5) = 0

⇒ r = or 2 5

5 2

Case i: Considering r = 5

2
, we get terms of G.P. are as 5

2
, 1,
2

Case ii: When r = 5

2
5
terms of G.P. are 2

5
, 1,
2
.
3+n
61. ∵ an = 4

3+1
a1 =
4
=1
3+2 5
a2 = =
4 4
3+3 6 3
a3 = = =
4 4 2

Now,
5−4
a2 - a1 = 5

4
− 1 =
4
=
1

3 5 6−5 1
a3 - a2 = 2
−
4
=
4
=
4

∵ a3 - a2 = a2 - a1 = 1

3+n
∴ The sequence defined by 4
is an A.P in which common difference is 1

4
.​​
62. Let a be the first term and r the common ratio and
T4 = 16, T7 = 128 (given)

∴ ar4 - 1 = 16 and ar7 - 1 = 128

ar3 = 16 ...(i)
ar6 = 128 ...(ii)
Dividing eq. (iii) by (i),
6
ar 128
=
3 16
ar

∴ r3 =
8 = 23
∴ r = 2

ar3 = 16 ⇒ a × 8 = 160
∴ a=2
Hence, first term is 2 and the common ratio is 2.
63. Given series be 1 + 2 + 4 ... + 512.
Here, a = 1, r = 2, and let Tn = arn - 1 = 512
n
a( r −1)
∵ Sn = r−1
as r > 1
n−1
ar r−a
S6 = r−1

512×2−1
= 2−1

= 1024 - 1 = 1023
64. a1 = first term = 1

a, ar, ar2,... arn-1 be a G.P.

ar2 + a⋅ r4 = 90
⇒ ar2(1 + r2) = 90
⇒ 1⋅ r2(1 + r2) = 90 [∵ Let r2 = P]
⇒ P(1 + P) = 90

22 / 27
 ⇒ P + P2 = 90
⇒ P2 + P - 90 = 0
⇒ P2 + 10P - 9P - 90 = 0
⇒ P(P + 10) - 9(P + 10) = 0
(P - 9)(P + 10) = 0
P=9
P = -10 (Not define)
P = r2 = 9
r = ±3
r = common ratio = ± 3​​
65. Given, a5 + a9 = 79
(a + 4d) + (a + 8d) = 72
⇒ 2a + 12d = 72

2(a + 6d) = 72
a + 6d = 36 ...(i)
a7 + a12 = 97
⇒ (a + 6d) + (a + 11d) = 97
⇒ 2a + 17d = 97 ...(ii)
By solving equation (i) and (ii) we get
2(a + 6d) = (36) × 2

∴ d=5
Put the value of d in equation (i)
a + 6d = 36
⇒ a + 30 = 36

∴ a = 36 - 30 = 6
a, a + d, a + 2d, ..., a + (n - 1)d are in A.P.
Hence, 6, 11, 16, ..., is an A.P.​​
66. Let three angles a - d, a, a + d are in an A.P.
a - d + a + a + d = 180o [We know that sum of all angles of a triangle is 180o]
⇒ 3a = 180o
a = 60o
Again, a + d = 2(a - d) [∵ a = 60o]
⇒ 60 + d = 2(60 - d)

⇒ 60 + d = 120 - d

⇒ 60 - 120 = -2d - d

⇒ -60 = -3d
−60
∴ d = = 20
−3

60 - 20, 60, 60 + 20 are all three angles of a triangle.

40, 60, 80 are in an A.P.​​
67. Given AP is 20, 19 , 18 , … 1

3
2

Here, a = 20
58 58−60 −2
and d = 19 1

3
− 20 =
3
− 20 =
3
= 3

Let n terms of the given AP be required to get the sum 300.

We know that, S = [2a + (n − 1)d]
n
n

−2
⇒ 300 = n

2
[2(20) + (n − 1) (
3
)] [∵ a = 20 and d = - 2/3]

⇒ 600 = n (40 − 2

3
n+
2

3
)

1 2
∴ 600 = (120n − 2n + 2n)
3

23 / 27
 2
⇒ 600 × 3 = 122n − 2n

⇒ 1800 + 2n2 - 122n = 0

⇒ 2(n2 - 61n + 900) = 0
⇒ n2 - 61n + 900 = 0 [∵ 2 ≠ 0]
⇒ n2 - 36n - 25n + 900 = 0
⇒ n(n - 36) - 25 (n - 36) = 0

⇒ (n - 36)(n - 25) = 0

⇒ n = 36 or 25

Since, a is positive and d is negative, so both values of n are possible. Hence, sum of first 25 terms of given AP
Hence, sum of first 25 terms of given AP = Sum of first 36 terms of given AP = 300.
–
68. 3
+ + √5 + ... to 25 terms
4

√5 √5

a1 =
3
,a 2 =
4

√5 √5

4 3 4−3 1
d = a2 - a1 = − = =
√5 √5 √5 √5

– 5−8
d = a3 - a2 = √5 − 4
= =
1

√5 √5 √5

a1, a2, a3,..., a25 is an A.P.

an = a1 + (n - 1)d
3 1 1 3
⇒ a25 = + (25 − 1) [∵ d = , a1 = ]
√5 √5 √5 √5

3 24
= +
√5 √5

3+24 27
= =
√5 √5

n
Sn = [a1 + an ]
2

= n

2
[first term + last term]
25 3 27
= [ + ]
2 √5 √5

25 30
= ×
2 √5
–
= 75√5
–
∴ S25 = 75√5
–
Sum of the 25 terms of an A.P is 75√5.

69. Given, a2 = a⋅ rn-1

−1
⇒
2
= a. r ...(i)
a5 = a⋅ r5-1
⇒
1

16
= a⋅ r
4
...(ii)
Eqn (ii) ÷ (i)
1
4
16 a⋅r
=
1 a⋅r
−
2

3 1 2
⇒ r = − ×
16 1

3 1
r = −
8

∴ r=− 1

Put the value of r in equation (i)

a⋅ r = − 1

−1 −1
⇒ a× ( ) =
2 2

∴ a=1
n
a(1− r )
S8 =
1−r

1 8
1[1− (− ) ]
2 (256−1)

−1
=
256
×
2

3
= 85

128
1−( )
2

85
Hence, the sum of 8 terms is 128
.​​
70. Given two A.P 9, 7, 5, ... and 24, 21, 18 ...
For first A.P, a = 9

24 / 27
 d = -2
an = a + (n - 1)d
= 9 + (n - 1)(-2)
= 9 - 2n + 2 = 11 - 2n
For 2nd A.P
a = 24
d = -3
an = a + (n - 1)d
= 24 + (n - 1)(-3) = 24 - 3n + 3 = 27 - 3n
According to question
11 - 2n = 27 - 3n
⇒ -2n + 3n = 27 - 11

n = 16
The value of n is 16.
Also, a16 = 11 - 2n = 11 - 2 × 16 = 11 - 32 = -21
a16 = 27 - 3n = 27 - 3 × 16 = 27 - 48 = -21​​
71. Given:
a = 5, d = 3, tn = 80, n = ?
∵ tn = a + (n - 1)d
80 = 5 + (n - 1)d
80 = 5 + 3n - 3
78 = 3n
n = 26
Hence, no. of terms of the given AP = 26​​
72. Simply by Putting n = 1, 2, 3 and 4 we get the Ist four terms.
i. an = 2n + 3
a1 = 2(1) + 3 = 5
a2 = 2(2) + 3 = 7
a3 = 2(3) + 3 = 9
a4 = 2(4) + 3 = 11
Hence, first four terms of above sequence are 5, 7, 9 and 11.
ii. an = (-1)n-1

a1 = (-1)1-1 = 1

a2 = (-1)2-1 = -1

a3 = (-1)3-1 = 1

a4 = (-1)4-1 = -1
Hence, 1st four terms of the above sequence are 1, -1, 1 and -1.
2
(n−1)
iii. an =
3
2
(1−1)
a1 = = 0
3
2
(2−1)
1
a2 = =
3 3
2
(3−1)
4
a3 = =
3 3
2
(4−1)
9
a4 = = = 3
3 3

Hence, Ist four terms of the above sequence ale 0, 1

3
, 4

3
and 3.​​
73. i. 6, 12, 18,...
a = 6, d = 12 - 6 = 6
a4 = a + 3d = 6 + 3(6) = 6 + 18 = 27

25 / 27
 a5 = a + 4d = 6 + 4(6) = 6 + 24 = 30
∴ a5 - a4 = 30 - 24
a5 - a4 = 6
1 4 7
ii. 3
,
3
,
3
...

1
a=
3
4 1
d = −
3 3
4−1 3
⇒ = = 3
3 1
1+27
a4 = a + 3d = 1

3
+ 3(3) =
1

3
+ 9 = 3
=
28

1+36
a5 = a + 4d = 1

3
+ 4(3) =
1

3
+ 12 = 3
−
37

37 28 37−28 9
∴ a5 - a4 = 3
−
3
= 3
=
3
=3
a5 - a4 = 3
iii. a = -15
d = -13 - (-15) = -13 + 15 = 2
∴ a4 = a + 3d = -15 + 3(2) = -15 + 6 = -9
a5 = a + 4d = -15 + 4(2) = -15 + 8 = -7
∴ a5 - a4 = -7 - (-9) = -7 + 9 = 2
a5 - a4 = 2​​

74. a, ar, ar2, ..., a⋅ rn-1 be a G.P.

a8 = a⋅ r8-1 [r = common ratio = 2]

⇒ 192 = 9.27
192 3
∴ a= =
7 2
2

a = first term = 3

a12 = a⋅ r12-1
3 11
= ⋅ (2)
2

= 3 × 211-1
= 3 × 210 = 3072
Hence, 12th term of a G.P is 3072.​​
75. Given 8, 4, 2,... are in G.P. 1024
1

a=8
r= 4

8
=
1

an = a⋅ rn-1
1 1 n−1
⇒ = 8 × ( )
1024 2

⇒ 2-10 = 2 3
(2)
1−n

⇒ 2-10 = 23+(1-n) [∵ if am = an then m = n]

⇒ 24-n = 2-10
⇒ 4 - n = -10

⇒ -n = -10 - 4

∴ n = 14

no. of terms is 14.

From the end of the series
a= ,r=2
1024
1

a6 = a⋅ r6-1

=
1024
1
⋅ 2
5
= 2
−10
× 2
5
= 2-10+5 = 2-5 = 1

32

Hence, the 6th term from end is 1

32

Again, a4 = a⋅ r4-1

26 / 27
=
1

1024
3
2
−10
= 2
3
× 2 = 2-7 = 1

128

Hence, the 4th term from end is 1

128
​

27 / 27