The 4th, 6th and last term of a geometric progression are 10, 40 and 640 respectively. If the common ratio is positive, find the first term, common ratio and the number of terms of the series.
G.P.
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Question

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

KnowledgeBoat · Accepted Answer

We know that, nth term of a G.P. is given by, Tn = arn - 1 Given, 4th term = 10 ⇒ T4 = ar3 ⇒ ar3 = 10 .....(1) Given, 6th term = 40 ⇒ T6 = ar5 ⇒ ar5 = 40 .....(2) Given, Let nth term be last term. ⇒ Tn = arn - 1 Since, last term = 640 ⇒ arn - 1 = 640 .....(3) Divide Equation 2 by Equation 1: ⇒ r5 - 3 = 4 ⇒ r2 = 4 ⇒ r = 2 (Since, common ratio is positive) Substitute r = 2 into Equation 1: ⇒ a(2)3 = 10 ⇒ a(8) = 10 ⇒ a = ⇒ a = . Now, substituting values of a and r in equation (3), we get : ⇒ 2n - 1 = 640 ⇒ 2n - 1 = ⇒ 2n - 1 = 128 × 4 ⇒ 2n - 1 = 512 ⇒ 2n - 1 = 29 ⇒ n - 1 = 9 ⇒ n = 9 + 1 ⇒ n = 10. Hence, a = , r = 2 and n = 10.

Mathematics

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

G.P.

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