Mathematics · Sequence and Series

JEE Advanced 2020 — Paper 1 — Question 41

Let $a_{1}, a_{2}, a_{3}, \ldots .$ . be a sequence of positive integers in arithmetic progression with common difference 2 .

Also, let $b_{1}, b_{2}, b_{3}, \ldots .$ . be a sequence of positive integers in geometric progression with common ratio 2. If

$\mathrm{a}_{1}=\mathrm{b}_{1}=\mathrm{c}$ , then the number of all possible values of c , for which the equality

$2\left(a_{1}+a_{2}+\ldots . .+a_{n}\right)=b_{1}+b_{2}+\ldots . .+b_{n}$ holds for some positive integer $n$ , is $\_\_\_\_$

Answer: 1

Numerical answer — enter this value.

$2\left(a_{1}+a_{2}+\ldots . .+a_{n}\right)=b_{1}+b_{2}+\ldots . .+b_{n}$

$\Rightarrow 2\left(\frac{\mathrm{n}}{2}(2 \mathrm{c}+(\mathrm{n}-1) 2)\right)=\mathrm{c}\left(\frac{2^{\mathrm{n}}-1}{2-1}\right)$

$c=\frac{2 n(n-1)}{2^{n}-1-2 n}$

$2^{x-2}>x^{2}-x \quad$ if $x \geq 8$

$2^{x}-1>2 x \quad$ if $x>3$

So, $\mathrm{c}<\frac{2^{\mathrm{n}-1}}{2^{\mathrm{n}}-1} \forall \mathrm{n} \geq 8$

as $\mathrm{c} \geq 1$ values of c are not possible for $\mathrm{n} \geq 8$

possible values of $n=3,4,5,6,7$ ; only satisfy when $n=3$ and $c=12$

Answer key and solution verified before publishing.

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