The second term of a geometric progression is 3. If its sum to infinity is \(\frac{25}{2}\), find the value of its common ratio.
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Correct Answer: Option n
Explanation:
\(T_{n} = ar^{n - 1}\) (Terms of a geometric progression)
\(T_{2} = ar = 3 ..... (1)\)
\(S_{\infty} = \frac{a}{1 - r}\) (Sum to infinity of a GP)
\(\frac{a}{1 - r} = \frac{25}{2} \implies 2a = 25(1 - r)\)
\(a = \frac{25 - 25r}{2}\)
\(3 = (\frac{25 - 25r}{2})r\)
\(6 = 25r - 25r^{2}\)
\(25r^{2} - 25r + 6 = 0\)
\(25r^{2} - 15r - 10r + 6 = 0\)
\(5r(5r - 3) - 2(5r - 3) = 0 \implies (5r - 3)(5r - 2) = 0\)
\(r = \frac{2}{5} \text{or r = } \frac{3}{5}\).
\(T_{n} = ar^{n - 1}\) (Terms of a geometric progression)
\(T_{2} = ar = 3 ..... (1)\)
\(S_{\infty} = \frac{a}{1 - r}\) (Sum to infinity of a GP)
\(\frac{a}{1 - r} = \frac{25}{2} \implies 2a = 25(1 - r)\)
\(a = \frac{25 - 25r}{2}\)
\(3 = (\frac{25 - 25r}{2})r\)
\(6 = 25r - 25r^{2}\)
\(25r^{2} - 25r + 6 = 0\)
\(25r^{2} - 15r - 10r + 6 = 0\)
\(5r(5r - 3) - 2(5r - 3) = 0 \implies (5r - 3)(5r - 2) = 0\)
\(r = \frac{2}{5} \text{or r = } \frac{3}{5}\).