Question
If the $(m+1)^{\text {th }},(n+1)^{\text {sh }}$ and $(r+1)^{t h}$ terms of an A.P. are in G.P., $m, n, r$ are in H.P. show that the ratio of the common difference to the first term in the A.P. is $-\frac{2}{n}$.
Step 1
P., we have $\frac{2}{n} = \frac{1}{m} + \frac{1}{r}$. This can be rewritten as $2mr = m + r$. Show more…
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If the $(m+1)$ th, $(n+1)$ th and $(r+1)$ th terms of an A.P. are in G.P. and $m, n, r$ are in H.P., then the ratio of the first term of the A.P. to its common difference is (A) $\frac{n}{3}$ (B) $-\frac{n}{3}$ (C) $\frac{n}{2}$ (D) $-\frac{n}{2}$
If $S_{n}$ represents the sum of $n$ terms of a G.P. whose first term and common ratio are $a$ and $r$ respectively, then prove that i. $\quad S_{1}+S_{2}+S_{3}+\ldots+S_{n}=\frac{n a}{1-r}-\frac{\operatorname{ar}\left(1-r^{n}\right)}{(1-r)^{2}}$; ii. $S_{1}+S_{3}+S_{5}+\ldots+S_{2 n-1}=\frac{a n}{1-r}-\frac{a r\left(1-r^{2 n}\right)}{(1-r)^{2}(1+r)}$.
If the ratio of the sum of $m$ term and $n$ terms of an A.P. be $m^{2}: n^{2}$, prove that the ratio of its $m$ th and $n$ th terms will be $2 m-1: 2 n-1$.
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