The sums of first n terms of two A.P.'s are in the ratio (7n + 2) :


The sums of first n terms of two A.P.'s are in the ratio (7n + 2) : (n + 4). Find the ratio of their 5th terms.


Let the first term, the common difference and the sum of the first $n$ terms of the first A .P. be $a_{1}, d_{1}$ and $\mathrm{S}_{1}$, respectively, and those of the second A.P. be $a_{2}, d_{2}$ and $\mathrm{S}_{2}$, respectively.

Then, we have,

$S_{1}=\frac{n}{2}\left[2 a_{1}+(n-1) d_{1}\right]$

And, $S_{2}=\frac{n}{2}\left[2 a_{2}+(n-1) d_{2}\right]$


$\frac{S_{1}}{S_{2}}=\frac{\frac{n}{2}\left[2 a_{1}+(n-1) d_{1}\right]}{\frac{n}{2}\left[2 a_{2}+(n-1) d_{2}\right]}=\frac{7 n+2}{n+4}$

$\Rightarrow \frac{S_{1}}{S_{2}}=\frac{\left[2 a_{1}+(n-1) d_{1}\right]}{\left[2 a_{2}+(n-1) d_{2}\right]}=\frac{7 n+2}{n+4}$

To find the ratio of the 5 th terms of the two A.P.s, we replace $n$ by $(2 \times 5-1=9)$ in the above equation:

$\Rightarrow \frac{\left[2 a_{1}+(9-1) d_{1}\right]}{\left[2 a_{2}+(9-1) d_{2}\right]}=\frac{7 \times 9+2}{9+4}$

$\Rightarrow \frac{\left[2 a_{1}+(8) d_{1}\right]}{\left[2 a_{2}+(8) d_{2}\right]}=\frac{7 \times 9+2}{9+4}=\frac{65}{13}$

$\Rightarrow \frac{\left[a_{1}+4 d_{1}\right]}{\left[a_{2}+4 d_{2}\right]}=\frac{5}{1}=5: 1$

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