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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. | |