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The sum pf 'n' terms of two arithmentic ...

The sum pf 'n' terms of two arithmentic progressions are in the ratio `(3n+8):(7n+15)`. Find the ratio of their `12^(th)` terms .

Text Solution

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`("sum to'n'terms of first.A.P")/("sum to 'n' terms of second A.P")=(3n+8)/(7n+15)`
`((n)/(2)[2a+(n-1)d])/((n)/(2)[2A+(n-1)D])=(3n+8)/(7n+15)rArr(2a+(n-1)d)/(2A+(n-1)D)=(3n+8)/(7n+15)`
Now `(12^(th)"term of "1^(st)"A.P")/(12^(th)"term of " 2^(nd) "A.P")=(a+11d)/(A+11D)`=?
Put n = 23 in (1)
`(2a+22d)/(2A+22D)=(3(23)+8)/(7(23)+15)rArr(a+11d)/(A+11D)=(7)/(16)`
`therefore` Required ratio `=7:16`
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