If the ratio of the sum of n terms of two A.P | vedclass

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Question

(c) Let ${S_n}$ and $S{'_n}$ be the sums of $n$ terms of two $A.P.'s$ and

${T_{11}}$ and $T{'_{11}}$ be the respective ${11^{th}}$ term

Vedclass · Accepted Answer

$4:3$

Vedclass · Answer

(c) Let ${S_n}$ and $S{'_n}$ be the sums of $n$ terms of two $A.P.'s$ and

${T_{11}}$ and $T{'_{11}}$ be the respective ${11^{th}}$ terms, then

$\frac{{{S_n}}}{{S{'_n}}} = \frac{{\frac{n}{2}[2a + (n - 1)d]}}{{\frac{n}{2}[2a' + (n - 1)d']}} = \frac{{7n + 1}}{{4n + 27}}$

$ \Rightarrow $ $\frac{{a + \frac{{(n - 1)}}{2}d}}{{a' + \frac{{(n - 1)}}{2}d'}} = \frac{{7n + 1}}{{4n + 27}}$

Now put $n = 21$,

we get $\frac{{a + 10d}}{{a' + 10d'}} = \frac{{{T_{11}}}}{{T{'_{11}}}} = \frac{{148}}{{111}} = \frac{4}{3}$.

Note : If ratio of sum of $n$ terms of two $A.P.'s$ are given in terms of $n$ and ratio of their ${p^{th}}$ terms are to be found then put $n = 2p - 1$.

Here we put $n = 11 	imes 2 - 1 = 21$.

If the ratio of the sum of $n$ terms of two $A.P.'s$ be $(7n + 1):(4n + 27)$, then the ratio of their ${11^{th}}$ terms will be

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