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Mathematics

How many terms of the G.P. 3,32,34,....3, \dfrac{3}{2}, \dfrac{3}{4}, …. are needed to give the sum 3069512?\dfrac{3069}{512}?

AP GP

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Answer

Here, a = 3 and r = 323=32×3=12.\dfrac{\dfrac{3}{2}}{3} = \dfrac{3}{2 \times 3} = \dfrac{1}{2}.

Let n terms are needed for the sum.

Sn=a(rn−1)r−1⇒3069512=3[(12)n−1]12−1⇒3069512=3[(12)n−1]1−22⇒3069512=−6[(12)n−1]⇒3069512×6=1−(12)n⇒(12)n=1−30693072⇒(12)n=3072−30693072⇒(12)n=33072⇒(12)n=11024⇒(12)n=(12)10∴n=10.S_n = \dfrac{a(r^n - 1)}{r - 1} \\[1em] \Rightarrow \dfrac{3069}{512} = \dfrac{3\Big[\Big(\dfrac{1}{2}\Big)^n - 1\Big]}{\dfrac{1}{2} - 1} \\[1em] \Rightarrow \dfrac{3069}{512} = \dfrac{3\Big[\Big(\dfrac{1}{2}\Big)^n - 1\Big]}{\dfrac{1 - 2}{2}} \\[1em] \Rightarrow \dfrac{3069}{512} = -6\Big[\Big(\dfrac{1}{2}\Big)^n - 1\Big] \\[1em] \Rightarrow \dfrac{3069}{512 \times 6} = 1 - \Big(\dfrac{1}{2}\Big)^n \\[1em] \Rightarrow \Big(\dfrac{1}{2}\Big)^n = 1 - \dfrac{3069}{3072} \\[1em] \Rightarrow \Big(\dfrac{1}{2}\Big)^n = \dfrac{3072 - 3069}{3072} \\[1em] \Rightarrow \Big(\dfrac{1}{2}\Big)^n = \dfrac{3}{3072} \\[1em] \Rightarrow \Big(\dfrac{1}{2}\Big)^n = \dfrac{1}{1024} \\[1em] \Rightarrow \Big(\dfrac{1}{2}\Big)^n = \Big(\dfrac{1}{2}\Big)^{10} \\[1em] \therefore n = 10.

Hence, 10 terms of the G.P. are required for the sum of 3069512\dfrac{3069}{512}.

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