Question

In a G.P sum of 2™, 3° and 4™ terms is 3 and that of 6, 7" and 8" term is 243 then S₅₀ =

Answer 1

ANSWER.

$\sf \to \: s_{50} \: = \dfrac{3 {}^{50} - 1}{26} \\ \\ \sf \to \: option \: (c) \: is \: correct$

EXPLANATION.

$\sf \to \: in \: a \: g.p \\ \\ \sf \to \: sum \: \: of \: \: 2nd, \: 3rd \: and \: 4th \: term \: = 3....(1) \\ \\ \sf \to \: the \: sum \: of \: 6th ,\: 7th \: and \: 8th \: term \: = 243$

To find the 50th term of g.p,

$\sf \to \: nth \: term \: of \: an \: g.p \: = ar {}^{n - 1} \\ \\ \sf \to \: ar + ar {}^{2} + ar {}^{3} = 3 \\ \\ \sf \to \: ar(1 + r + {r}^{2} ) = 3 \\ \\ \sf \to \: (1 + r + {r}^{2} ) = \dfrac{3}{ar} ......(1)$

$\sf \to \: ar {}^{5} + {ar}^{6} + {ar}^{7} = 243 \\ \\ \sf \to \: ar {}^{5} (1 + r + {r}^{2} ) = 243 ......(2)$

$\sf \to \: put \: the \: value \: of \: equation \: (1) \: in \: (2) \\ \\ \sf \to \: ar {}^{5} \times ( \frac{3}{ar} ) = 243 \\ \\ \sf \to \: {r}^{4} \times 3 = 243 \\ \\ \sf \to \: {r}^{4} = 81 \\ \\ \sf \to \: {r}^{4} = (3) {}^{4} \\ \\ \sf \to \: r \: = 3$

$\sf \to \: put \: r \: = 3\: in \: equation \: (1) \\ \\ \sf \to \: a \times 3(1 + 3 + 9) = 3 \\ \\ \sf \to \: a \: = \dfrac{1}{13}$

$\sf \to \: sum \: of \: n \: terms \: of \: g.p \\ \\ \sf \to \: s_{n} \: = \dfrac{a( {r}^{n} - 1)}{r - 1} \\ \\ \sf \to \: s_{50} \: = \dfrac{1}{13} \times \frac{(3 {}^{50} - 1)}{3 - 1} \\ \\ \sf \to \: s_{50} \: = \frac{3 {}^{50} - 1}{26} = answer$

Answer 2

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In a G.P sum of 2™, 3° and 4™ terms is 3 and that of 6, 7" and 8" term is 243 then S₅₀ =