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110, (be + ca + ab)’ = Dbée? + Jabe (a +b +.)
= ple sa)re le st)ee(o'+e)}
+ abe (a+ bre) -0
We have, acy as
2
lapplying AM > GM for the pairs (c 3°}; (@,b’)s
(b3,c*)]
Substituting in (1)
Gercaraby?> E [bx 2ea 4 ey 20b +58 2be]
4 dabe (a+b +0) > ; x Dabe (a + b+ €) + 2abe
(a+b+0)>dabe(arb+o)
111. Statement 2 is the prime factorization theorem.
Ale pj py =p, be the prime numbers in AP
‘Their common difference =
‘Then the (p, + L)th term of this progression is
ta 8a composite number whichis contradiction
Statement | is true but Statement 1 does not fol-
low from Statement 2
=> Choice (b)
112, Statement 2s standard result
Statement 2: §,
rt
F+T
=> §.-r=1 which does not depend on r
Statement Lis false
> Choice (d)
113. Statement 2 isa standard result
a + 2bc, b? + 2ac, + 2ab are in AP
=> ab+2be—k,b'+ 2ac—k, c+ 2ab— kin AP
(k=ab+be+ ca)
= ~(a- bea), 40 - Ofa—b), e-Nb-)
in AP
(2=bYfe-2) (blob) (e=a(o—
(where m = ~(a—b)(b ~€) (e-a)
a
> scecpipinar
= b-oe-aa-bin HP
Statement 1 is true and follows from Statement 2
=> Choice (a)
Sequences and Series 5.71
114. After the first halflife, the remaining sample is 3
After the next half life the remaining sample is
Ifx)_x
ili)
Similarly after the third half cycle the remaining
sample 2 andsoon
Fin 0,2, 000
=
118. 12 years= 6 half cycles
‘quantity let over alter 6 half cycles
1
= 320045 = 50g
116. 12.596 of isotope = i of the original
~E itil aurind with xual
8 ?
half-life= : years
After 8 more years total period = 8 + 8= 16 years
‘a I 00 _ 25
quantity rematning = So = P=
117, We observe thatthe last term in the nth row
= nth term of the sequence 1°, 3°, 6 10
= square of the nth term of the sequence 1, 3,
6.1000
ica
ie
(a) istrue
nth row has n terms of the term.
(x41 429, em
(ort) nlo=t)
We know, x+n=
a(n-1)
<. Firstterm= +1] => (bc) are true
Now
'S, = Sum of the elements in the nth row
¥(x+ny’
