JEE Advanced 2016 · Paper 2 · Mathematics Q38

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MathematicsSingle correct+3 / -1Question 38 of 54

Let $b_i>1$ for $i=1,2,\ldots,101$ . Suppose $\log_eb_1,\log_eb_2,\ldots,\log_eb_{101}$ are in Arithmetic Progression (A.P.) with the common difference $\log_e2$ . Suppose $a_1,a_2,\ldots,a_{101}$ are in A.P. such that $a_1=b_1$ and $a_{51}=b_{51}$ . If $t=b_1+b_2+\cdots+b_{51}$ and $s=a_1+a_2+\cdots+a_{51}$ , then

s>t

and

a_{101}>b_{101}

s>t

and

a_{101}<b_{101}

s<t

and

a_{101}>b_{101}

s<t

and

a_{101}<b_{101}

Answer and worked solution

Official answer: B

The sequences. The $b_i$ form a GP with ratio 2. The $a_i$ form an AP with the same first and 51st terms.

Sums. Between the shared ends, the AP lies above the convex GP: $a_i>b_i$ for $1<i<51$ . So $s>t$ .

101st terms.
$a_{101}=2a_{51}-a_1=b_1(2^{51}-1)<b_1\cdot2^{100}=b_{101}.$

(B)

Topics:AM-GM-HM inequalities Geometric progression Arithmetic progression

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JEE Advanced 2016 · Paper 2, Mathematics question 38