Question

the 4th, 6th and the last term of a gp are 10,40 and 640 respectively. If the common ratio is positive, find the first term, common ratio and the number of terms in the series.​

Answer 1

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Answer 2

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

• we now the general term of G.P:

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

$4^{th} \: term = 10\: (given)$

$\implies a r^{3} = 10 \: ---(1)$

$and \: 6^{th} \: term = 40\: (given)$

$\implies a r^{5} = 40 \: ---(2)$

/* Do Equation (2) ÷ equation (1) */

$\implies \frac{ar^{5}}{ar^{3}} = \frac{40}{10}$

$\implies r^{2} = 4$

$\implies r = ±\sqrt{2^{2}}$

$\implies r = ±2$

/* It is given that common ratio is positive */

Therefore. ,

$\implies r = 2 \: ---(3)$

/* Substitute r = 2 in the equation (1) , we get */

$\implies a\times 2^{3} = 10$

$\implies a = \frac{10}{8}$

$\implies a = \frac{5}{4} \: ---(4)$

$\red { First\: term (a) } \green { = \frac{5}{4}}$

$\red { Common \: ratio (r)} \green { = 2}$

$Now , a_{n} = 640 \: (given)$

$\implies ar^{n-1} = 640$

$\implies \frac{5}{4} \times 2^{n-1} = 640$

$\implies 2^{n-1} = 640 \times \frac{4}{5}$

$\implies 2^{n-1} = 128 \times 4$

$\implies 2^{n-1} = 2^{9}$

$\implies n - 1 = 9$

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

$\implies n = 10$

Therefore.,

$\red { First\: term (a) } \green { = \frac{5}{4}}$

$\red { Common \: ratio (r)} \green { = 2}$

$\red { Number \: of \:terms \: in \:G.P} \green {= 10}$

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