Question

find the sum of first 10 terms of a GP whose 3rd term is 12 and it's 8th term is 384​

Answer 1

$Let \: a \: and \: r \:are \: first \: term \: and \\common \:ratio \:of \: given \: G.P.$

$\boxed { \pink { n^{th} \:term (a_{n}) = a r^{n-1} }}$

$Given \: 3^{rd} \:term (a_{3}) = 12$

$\implies ar^{2} = 12 \: ---(1)$

$Given \: 8^{th} \:term (a_{8}) = 384$

$\implies ar^{7} = 384 \: ---(2)$

/* Divide equation (2) by equation (1) , we get */

$\frac{ar^{7}}{ar^{2}} = \frac{384}{12}$

$\implies r^{5} = 32$

$\implies r^{5} = 2^{5}$

$\implies r = 2$

$\boxed { \pink { Since, If \: a^{m} = a^{n} \implies m = n }}$

/* Substitute r = 2 in equation (1) */

$a\times 2^{2} = 12$

$\implies a = \frac{12}{4} = 3$

$\boxed { \pink { Sum \:of \: n \: terms (S_{n}) = \frac{a(r^{n}- 1)}{(r-1) } }}$

$Here, a = 3 , r = 2 , \: and \: n = 10$

$S_{10} = \frac{3( 2^{10} - 1)}{(2-1)}\\= \frac{3( 1024 -1 )}{1}\\= 3 \times 1023\\= 3069$

Therefore.,

$\red { Sum \: of \: 10 \:terms }\green { = 3069}$

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