Question

The first term of G.P is 2 and the sum to infinity is 6.Find the common ratio

Answer 1

Answer:

$\frac{2}{3}$

Step-by-step explanation:

$Given \: \: \\ first \: term \: of \: GP \: = 2 \\ S \infty = 6 \\ We \: know \: that \: S \infty = \frac{a}{1 - r} \\ Here \: r \: is \: common \: ratio \: and \: a \: is \: the \: first \: term. \\ According \: to \: the \: problem \\ 6 = \frac{2}{1 - r} \\ 6(1 - r) = 2 \\ 3(1 - r) = 1 \\ 3 - 3r = 1 \\ 3 - 1 = 3r \\ 2 = 3r \\ r = \frac{2}{3} \\ Therefore \: common \: ratio \: in \: the \: given \: GP \: \\ = \frac{2}{3}$

Answer 2

$\huge \underline \bold \green{QUESTION}:$

The first term of G.P is 2 and the sum to infinity is 6 . Find the common ratio

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$\huge \underline \bold \red{SOLUTION}:$

$the \: sum \: of \: infinite \: terms \: = \frac{ a}{1 - r} \\ where \: a \: is \: first \: term \: and \: r \: is \: common \: ratio \\ in \: the \: given \: question \: \\ a = 2 \\ r = ? \: \\ s = 6 \\ s = \frac{a}{1 - r} \\ = > 6 = \frac{2}{1 - r} \\ = > 1 - r = \frac{2}{6} \\ = > 1 - r = \frac{1}{3} \\ = > r = 1 - \frac{1}{3} \\ = > r = \frac{2}{3}$

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$\underline\bold \red{final \: answer}:$

$\bold{r = \frac{2}{3} }$

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