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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
$2:3$
$3:4$
$4:3$
$5:6$
Solution
(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 $n = 11 \times 2 – 1 = 21$.
