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Standard XIII

Mathematics

Sum of Infinite Terms of a GP

Let an denote...

Question

Let $a_{n}$ denote the $n^{th}$ term of a G.P. If $a_{1} = 3, a_{n} = 96$ and sum of $n$ terms of the series is $189$ , then the value of $n$ is

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Solution

Let $r$ be the common ratio of the given G.P., so
$a_{1} = a a_{n} = 96 \Rightarrow a r^{n - 1} = 96 \Rightarrow 3 \cdot r^{n - 1} = 96 \Rightarrow r^{n - 1} = 32 \dots (i)$

Now, $S_{n} = 189$
$\Rightarrow a (\frac{r^{n} - 1}{r - 1}) = 189 \Rightarrow 3 (\frac{32 r - 1}{r - 1}) = 189 \Rightarrow \frac{32 r - 1}{r - 1} = 63 \Rightarrow 31 r = 62 \Rightarrow r = 2$
Using equation $(1)$ , we get
$2^{n - 1} = 32 ∴ n = 6$

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Sum of Infinite Terms of a GP

Standard XIII Mathematics

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Let an denote the nth term of a G.P. If a1=3,an=96 and sum of n terms of the series is 189, then the value of n is

Let r be the common ratio of the given G.P., so a1=aan=96⇒arn−1=96⇒3⋅rn−1=96⇒rn−1=32 ⋯(i) Now, Sn=189 ⇒a(rn−1r−1)=189⇒3(32r−1r−1)=189⇒32r−1r−1=63⇒31r=62⇒r=2 Using equation (1), we get 2n−1=32∴n=6

Let $a_{n}$ denote the $n^{th}$ term of a G.P. If $a_{1} = 3, a_{n} = 96$ and sum of $n$ terms of the series is $189$ , then the value of $n$ is