Sequences and Series 2 Question 7

8.

Let bi>1 for i=1,2,,101. Suppose logeb1,logeb2, ,logeb101 are in AP with the common difference loge2 . Suppose a1,a2,,a101 are in AP, such that a1=b1 and a51=b51. If t=b1+b2++b51 and s=a1+a2++a51, then

(2016 Adv.)

(a) s>t and a101>b101

(b) s>t and a101<b101

(c) s<t and a101>b101

(d) s<t and a101<b101

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Answer:

Correct Answer: 8. (b)

Solution:

  1. If logb1,logb2,,logb101 are in AP, with common difference loge2, then b1,b2,,b101 are in GP, with common ratio 2 .

b1=20b1,b2=21b1,b3=22b1,,b101=2100b1

Also, a1,a2,,a101 are in AP.

Given, a1=b1 and a51=b51

a1+50D=250b1a1+50D=250a1[a1=b1].

Now, t=b1+b2++b51

t=b1(2511)21

and s=a1+a2++a51

=512(2a1+50D)

t=a1(2511)[a1=b1]

or t=251a1a1<251a1

and s=512[a1+(a1+50D)] [from Eq. (ii)]

=512[a1+250a1]

=512a1+512250a1

s>251a1

From Eqs. (v) and (vi), we get s>t

Also, a101=a1+100D and b101=2100b1

a101=a1+100250a1a150 and b101=2100a1a101=a1+251a12a1=251a1a1a101<251a1 and b101>251a1b101>a101



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