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If the ratio of the sum of first n terms of two A.P’s is $( 7n\ +1)$: $( 4n\ +\ 27)$, find the ratio of their $m^{th}$ terms.
Given: The ratio of the sum of first n terms of two A.P’s is $( 7n\ +1)$: $( 4n\ +\ 27)$,
To do: To find the ratio of their $m^{th}$ terms.
Solution:
Let $a_{1} ,\ a_{2}$ be the first terms and $d_{1}$ , $d_{2}$ the common differences of the two given A.P.’s.
Thus,We have sum of all its terms $S_{n} =\frac{n}{2}[ 2a+( n-1) d]$
Let S and S' are the sum of n terms of the given A.P.
$\frac{S}{S'} =\frac{\frac{n}{2}[ 2a_{1} +( n-1) d_{1}]}{\frac{n}{2}[ 2a_{2} +( n-1) d_{2}]}$
It is given that $\frac{S}{S'} =\frac{7n+1}{4n+27}$
$\therefore \frac{\frac{n}{2}[ 2a_{1} +( n-1) d_{1}]}{\frac{n}{2}[ 2a_{2} +( n-1) d_{2}]} =\frac{7n+1}{4n+27}$
$\frac{2a_{1} +( n-1) d_{1}}{2a_{2} +( n-1) d_{2}} =\frac{7n+1}{4n+27} \ \ \ \ \ ...................( 1)$
To find the ratio of the $m^{th}$ terms of the two given A.P's, replace $n$ by $( 2m–1)$ in equation $( 1)$
$\frac{2a_{1} +( 2m-1-1) d_{1}}{2a_{2} +( 2m-1-1) d_{2}} =\frac{7( 2m-1) +1}{4( 2m-1) +27}$
$\frac{2a_{1} +( 2m-2) d_{1}}{2a_{2} +( 2m-2) d_{2}} =\frac{14m-7+1}{8m-4+27}$
$\frac{2a_{1} +2( m-1) d_{1}}{2a_{2} +2( m-1) d_{2}} =\frac{14m-6}{8m+23}$
$\frac{2[ a_{1} +( m-1) d_{1}]}{2[ a_{2} +( m-1) d_{2}]} =\frac{14m-6}{8m+23}$
$\frac{[ a_{1} +( m-1) d_{1}]}{[ a_{2} +( m-1) d_{2}]} =\frac{14m-6}{8m+23}$
$\because a_{m} =a+( m-1) d$
$\frac{m^{th\ } term\ of\ the\ First\ A.P.}{m^{th} \ term\ of\ second\ A.P.} =\frac{14m-6}{14m+20}$
Hence, The ratio of the terms of the two A.P's is $14m –6$:$8m+23$.
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