If the sum of infinite terms and the second term of a geometric progression is 16 : 3, then find the common ratio?
Explanation:
Let the first term of the GP be ‘a’ and the common ratio be ‘r’.
Given, S∞T2=163
⇒ a/(1-r)ar=163
⇒ 1r(1-r)=163
⇒ 16r2 - 16r + 3 = 0
⇒ r = ¼ or 3/4
Since only ¼ is there in one of the options and there is no option as ’Cannot be determined’, we will mark the correct answer as option (a).
Hence, option (a).
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