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Standard XII

Mathematics

Geometric Progression

The sum of ...

Question

The sum of $2 n$ terms of a geometric progression whose first term is $^{'} a^{'}$ and common ratio $^{'} r^{'}$ is equal to the sum of $n$ terms of a geometric progression whose first term is $^{'} b^{'}$ and common ' $r^{2}$ '. then $b$ is equal to

The sum of the first two terms of the first series.

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The sum of the first and last terms of the first series.

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The sum of the last two terms of the first series.

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None of these

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Solution

The correct option is A The sum of the first two terms of the first series.
Given that
$\begin{matrix} \frac{a (r^{2 n} - 1)}{r - 1} = \frac{b {(r^{2})}^{n} - 1}{r^{2} - 1} \Rightarrow \frac{a (r^{2 n} - 1)}{r - 1} = \frac{b (r^{2 n} - 1)}{(r - 1) (r + 1)} \Rightarrow b = a (r + 1) \Rightarrow b = a + a r \end{matrix}$
$b$ = sum of first two term of the first series.

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The sum of 2n terms of a geometric progression whose first term is ′a′ and common ratio ′r′ is equal to the sum of n terms of a geometric progression whose first term is ′b′ and common 'r2'. then b is equal to

The correct option is A The sum of the first two terms of the first series.Given thata(r2n−1)r−1=b(r2)n−1r2−1⇒a(r2n−1)r−1=b(r2n−1)(r−1)(r+1)⇒b=a(r+1)⇒b=a+arb= sum of first two term of the first series.

The sum of $2 n$ terms of a geometric progression whose first term is $^{'} a^{'}$ and common ratio $^{'} r^{'}$ is equal to the sum of $n$ terms of a geometric progression whose first term is $^{'} b^{'}$ and common ' $r^{2}$ '. then $b$ is equal to