Question

If the sum of first 6 terms of a g.p. is nine times of the sum of its first three terms, then its common ratio is

Answer 1

Let the first term be a and common ratio be r.
Given, S6 = 9 × S3
i.e.,
$\frac{a( {r}^{6} - 1) }{r - 1} = 9 \times \frac{a( {r}^{3} - 1) }{r - 1} \\ ( {r}^{3} + 1)( {r}^{3} - 1) = 9 \times ( {r}^{3} - 1) \\ {r}^{3} + 1 = 9 \\ {r}^{3} = 8 \\ common \: \: ratio \: \: \: r = 2$

Answer 2

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☞ Your Answer is 2

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$\huge\sf\blue{Given}$

✭ Sum of first 6 terms of a GP is nine times the sum of the first 3 terms

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◈ Common Difference?

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We know that,

$\sf S_6 = 9 × S_3$

That is,

$\sf\dashrightarrow\frac{a( {r}^{6} - 1) }{r - 1} = 9 \times \frac{a( {r}^{3} - 1) }{r - 1} \\\\\sf\dashrightarrow ( {r}^{3} + 1)( {r}^{3} - 1) = 9 \times ( {r}^{3} - 1) \\\\\sf\dashrightarrow {r}^{3} + 1 = 9 \\\\\sf\dashrightarrow {r}^{3} = 8 \\\\\sf\dashrightarrow Common \: \: Ratio \: \: \: R = 2$

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