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# If the ratio of the sum of the first n terms of two APs is $(7n + 1)$: $(4n + 27)$, then find the ratio of their 9$^{th}$ terms.

**Given:**The ratio of the sum of the first n terms of two APs is $(7n + 1)$:$(4n + 27)$

**To do:**To find the ratio of their 9$^{th}$ terms.

**Solution:**

Given two A.P.s with n terms each

For AP 1st,

let us say first term $a_{1}$ and common difference $d_{1}$

Then Sum of the n terms,

$S( 1st\ A.P.) =\frac{n}{2}( 2a_{1} +( n-1) d_{1})$

similarly first term of second A.P., $a_{2}$ and common difference $d_{2}$.

Sum of n terms,

$S( 2nd\ A.P.) =\frac{n}{2}( 2a_{2} +( n-1) d_{2})$

As given,

$\frac{S( 1st\ AP)}{S( 2nd\ AP)} =\frac{7n+1}{4n+27}$

$\Rightarrow \frac{\frac{n}{2}( 2a_{1} +( n-1) d_{1})}{\frac{n}{2}( 2a_{2} +( n-1) d_{2})} =\frac{7n+1}{4n+27}$

$\Rightarrow \frac{a_{1}+\frac{( n-1)}{2}d_{2}}{a_{2}+\frac{( n-1)}{2}d_{2}}=\frac{7n+1}{4n+27}$

Ratio of their $9^{th}$ terms$=\frac{a_{1}+8d_{1}}{a_{2}+8d_{2}}$

On comparing,

$\frac{a_{1}+\frac{( n-1)}{2}d_{2}}{a_{2}+\frac{( n-1)}{2}d_{2}}=\frac{a_{1}+8d_{1}}{a_{2}+8d_{2}}$

$\Rightarrow \frac{( n-1)}{2}=8$

$\Rightarrow n=17$

Substituting this value $n=17$,

Ratio of their $9^{th}$ term$=\frac{a_{1}+8d_{1}}{a_{2}+8d_{2}}=\frac{a_{1}+\frac{( n-1)}{2}d_{2}}{a_{2}+\frac{( n-1)}{2}d_{2}}=\frac{7n+1}{4n+27}=\frac{7( 17)+1}{4( 17)+27}=\frac{120}{95}$

Thus, The ratio $=24:19$.

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