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Q.
The sum of n terms of two arithmetic series are in the ratio $2n + 3 : 6n + 5$, then the ratio of their 13th terms is
Sequences and Series
Solution:
Sum of an A.P. is given by
$S_{n}=\frac{n}{2}\left[2a+\left(n-1\right)d\right]$
where 'a' is the first term and 'd' is the common difference of $A.P.$
Let $S_{n_1}$ be the sum of n terms of $I^{st}\, A.P.$
and $S_{n_2}$ be the sum of n terms of $II^{nd}\,A.P.$
Given that the sum of n terms of two arithmetic series is in the ratio $2n + 3 : 6n + 5$
$\Rightarrow \frac{S_{n_1}}{S_{n_2}}=\frac{2n+3}{6n+5}\,...\left(i\right)$
$\Rightarrow S_{n_1}=\frac{n}{2}\left[2a_{1}+\left(n-1\right)d_{1}\right]=2n+3$ and
$ S_{n_1}=\frac{n}{2}\left[2a_{2}+\left(n-1\right)d_{2}\right]=6n+5
$From Eq. (i) , we get
$\frac{S_{n_1}}{S_{n_1}}=\frac{\frac{n}{2}\left[2a_{1}+\left(n-1\right)d_{1}\right]}{\frac{n}{2}\left[2a_{2}+\left(n-1\right)d_{2}\right]}=\frac{2n+3}{6n+5}$
$\Rightarrow \frac{2a_{1}+\left(n-1\right)d_{1}}{2a_{2}+\left(n-1\right)d_{2}}=\frac{2n+3}{6n+5}$
For $a = 13, n = 2a - 1 = 2 × 13 - 1 = 25$
$\therefore \frac{2a_{1}+\left(25-1\right)d_{1}}{2a_{2}+\left(25-1\right)d_{2}}=\frac{53}{155} \Rightarrow \frac{a_{1}+12d_{1}}{a_{2}+12d_{2}}=\frac{53}{155}$