Question

JEE ADVANCED MATHS QUESTION SEPTEMBER 2020

Answer 1

ANSWER.

=> number of all possible values of c for which

the equality holds for a some positive integer

n = 1.

EXPLANATION.

$\sf : \implies \: a_{1} , a_{2} , a_{3} ....... \: be \: a \: sequence \: of \: positive \: integer \: in \: ap \\ \\ \sf : \implies \: common \: difference \: = 2 \\ \\ \sf : \implies \: b_{1} , b_{2} , b_{3}...... \: be \: a \: sequence \: of \: positive \: integer \: in \: gp \\ \\ \sf : \implies \: common \: ratio \: = 2 \\ \\ \sf : \implies \: a_{1} = b_{1} = c \: \: \: \: (given)$

$\sf : \implies \: equation \: = 2( a_{1} , a_{2} , a_{3}..... a_{n}) = b_{1} , b_{2} , b_{3}...... b_{n} \\ \\ \sf : \implies \: sum \: of \: n \: terms \: of \: an \: ap \\ \\ \sf : \implies \: s_{n} = \frac{n}{2}(2a \: + (n - 1)d) \\ \\ \sf : \implies \: sum \: of \: n \: terms \: of \: gp \\ \\ \sf : \implies \: s_{n} = \frac{a( {r}^{n} - 1)}{r - 1}$

$\sf : \implies \: 2 \times \dfrac{n}{2} (2c + (n - 1)2) = \dfrac{c( {2}^{n} - 1)}{2 - 1} \\ \\ \sf : \implies \: n(2c + (n - 1)2) = c( {2}^{n} - 1) \\ \\ \sf : \implies \: 2nc \: + n(n - 1)2 = c( {2}^{n} - 1) \\ \\ \sf : \implies \: c \: = \frac{2n(n - 1)}{ {2}^{n} - 2n - 1} \\ \\ \sf : \implies \: \: c > 1$

$\sf : \implies \: \dfrac{2n(n - 1)}{ {2}^{n} - 2n - 1} \geqslant 1 \\ \\ \sf : \implies \: 2(n)(n - 1) \geqslant \: {2}^{n} - 2n - 1 \\ \\ \sf : \implies \: 2n {}^{2} - 2n \geqslant \: {2}^{n} - 2n - 1 \\ \\ \sf : \implies \: {2n}^{2} \geqslant \: {2}^{n} - 1 \\ \\ \sf : \implies \: n \leqslant 6 \\ \\ \sf : \implies \: n \: = 1 \: \:only \: one \: \: values \:$