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If the ratio of sum of first ntg term of twi ap are 7n+1/4n+27.then find the ratio of its 9th term?

Answer» Let a1\xa0and a2 be the first terms and d1\xa0and d2 be the common difference of the two APs respectively.Let Sn and S\'n be the sums of the first n terms of the two APs and Tn and T\'n\xa0be their nth terms respectively.Then,\xa0{tex}\\frac { S _ { n } } { S _ { n } ^ { \\prime } } = \\frac { 7 n + 1 } { 4 n + 27 } \\Rightarrow \\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 + 1 } { 4 n + 27 }{/tex}{tex}\\Rightarrow \\frac { 2 a _ { 1 } + ( n - 1 ) d _ { 1 } } { 2 a _ { 2 } + ( n - 1 ) d _ { 2 } } = \\frac { 7 n + 1 } { 4 n + 27 }{/tex}\xa0........(i)To find the ratio of mth terms, we replace n by (2m -1) in the above expression.Replacing n by (2 {tex}\\times{/tex}\xa09 -1), i.e., 17 on both sides in (i), we get{tex}\\frac { 2 a _ { 1 } + ( 17 - 1 ) d _ { 1 } } { 2 a _ { 2 } + ( 17 - 1 ) d _ { 2 } } = \\frac { 7 \\times 17 + 1 } { 4 \\times 17 + 27 } \\Rightarrow \\frac { 2 a _ { 1 } + 16 d _ { 1 } } { 2 a _ { 2 } + 16 d _ { 2 } } = \\frac { 120 } { 95 }{/tex}{tex}\\Rightarrow \\frac { a _ { 1 } + 8 d _ { 1 } } { a _ { 2 } + 8 d _ { 2 } } = \\frac { 24 } { 19 }{/tex}{tex}\\Rightarrow \\frac { a _ { 1 } + ( 9 - 1 ) d _ { 1 } } { a _ { 2 } + ( 9 - 1 ) d _ { 2 } } = \\frac { 24 } { 19 }{/tex}{tex}\\Rightarrow \\frac { a _ { 1 } + ( 9 - 1 ) d _ { 1 } } { a _ { 2 } + ( 9 - 1 ) d _ { 2 } } = \\frac { 24 } { 19 }{/tex}{tex}\\Rightarrow \\frac { T _ { 9 } } { T _ { 9 } ^ { \\prime } } = \\frac { 24 } { 19 }{/tex}{tex}\\therefore{/tex}\xa0required ratio = 24:19.


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