Question

If the sum of infinite series 1+4x+7x^2+10x^3+.......=35/16 then x is​

Answer 1

EXPLANATION.

$\sf : \implies \: 1 + 4x + 7 {x}^{2} + 10 {x}^{3} ......... = \dfrac{35}{16}$

$\sf : \implies \: s_{ \infty } \: = 1 + 4x + 7 {x}^{2} + 10 {x}^{3} \: \: \: \: ......(1) \\ \\ \sf : \implies \: x s_{ \infty } \: = x + 4 {x}^{2} + 7 {x}^{3} + 10 {x}^{4} \: \: \: \: ......(2) \\ \\ \sf : \implies \: subtract \: equation \: \: (1) \: \: and \: \: (2) \\ \\ \sf : \implies \: (1 - x) s_{ \infty } \: = 1 + 3x + 3 {x}^{2} + 3 {x}^{3} ...... \infty$

$\sf : \implies \: first \: term \: = a \: = 3x \\ \\ \sf : \implies \: common \: ratio \: = r \: = x \\ \\ \sf : \implies \: sum \: of \: an \: infinite \: series \: of \: gp \\ \\ \sf : \implies \: s_{ \infty } \: = \frac{a}{1 - r} \: \: for \: \: |x| < 1$

$\sf : \implies \: (1 - x) s_{ \infty } \: = 1 + \dfrac{3x}{(1 - x)} \\ \\ \sf : \implies \: s_{ \infty } \: = \frac{35}{16} \\ \\ \sf : \implies \: (1 - x) \frac{35}{16} = \frac{1 + 2x}{1 - x} \\ \\ \sf : \implies \: 35(1 - x) {}^{2} = 16(1 + 2x)$

$\sf : \implies \: 35(1 + {x}^{2} - 2x) = 16 + 32x \\ \\ \sf : \implies \: 35 + 35 {x}^{2} - 70x = 16 + 32x \\ \\ \sf : \implies \: 35 {x}^{2} - 102x + 19 = 0 \\ \\ \sf : \implies \: 35 {x}^{2} - 95x - 7x + 19 = 0$

$\sf : \implies \: 5x(7x - 19) - 1(7x - 19) = 0 \\ \\ \sf : \implies \: (7x - 19)(5x - 1) = 0 \\ \\ \sf : \implies \: x \: = \frac{19}{7} \: \: \: and \: \: \: x = \frac{1}{5}$

$\sf : \implies \: x \: \ne \: \dfrac{19}{7} \: \: because \: s_{ \infty } \: = |x| < 1 \\ \\ \sf : \implies \: x \: = \frac{1}{5} = answer$

Answer 2

$S = 1 + 4x + 7 {x}^{2} + 10 {x}^{3} + ....... \infty = > AGP$

$xS = x + 4 {x}^{2} + 7 {x}^{3} + ....... \infty$

$\red{S(1 - n) = 1 + 3 + 3 {x}^{2} + 3 {x}^{3} + 3 {x}^{4} + ......... \infty }$

$\red{S(1 - 3) = 1 + 3x(1 + x - {x}^{2} + {x}^{3} ....... \infty )}$

$\red{ \frac{35}{16} (1 - x) = 1 - \frac{x + 3x}{1 - x} = \frac{1 + 2x}{1 - x} }$

$\red{35(1 - x {)}^{2} = 16(1 + 2x)}$

$\red{35(1 + {x}^{2} - 2x) = 16 + 32 x }$

$\red{35 {x}^{2} - 70x + 35 = 16 + 32x }$

$\boxed{ \color{lime} 35 {x}^{2} - {10}^{2} + 19 = 0 }$

$35 {x}^{2} - 95x + 7x + 19 = 0$

$5x(7x - 19x) - 1(7x - 19) = 0$

$n = \frac{1}{5} , \frac{19}{7}$