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.
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}).