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The ratio of the sum of n term of two AP is 7n+1:4n+24 find the ratio of their mth term

Answer» Let a, and A be the first terms and d and D be the common difference of two A.PsThen, according to the question,{tex}\\frac { S _ { n } } { S _ { n } ^ { \\prime } } = \\frac { \\frac { n } { 2 } [ 2 a + ( n - 1 ) d ] } { \\frac { n } { 2 } [ 2 A + ( n - 1 ) D ] } = \\frac { 7 n + 1 } { 4 n + 27 }{/tex}or,\xa0{tex}\\frac { 2 a + ( n - 1 ) d } { 2 A + ( n - 1 ) D } = \\frac { 7 n + 1 } { 4 n + 27 }{/tex}or,{tex}\\frac { a + \\left( \\frac { n - 1 } { 2 } \\right) d } { A + \\left( \\frac { n - 1 } { 2 } \\right) D } = \\frac { 7 n + 1 } { 4 n + 27 }{/tex}Putting,\xa0{tex}\\frac { n - 1 } { 2 } = m - 1{/tex}{tex}n-1 = 2m - 2{/tex}{tex}n= 2m - 2 + 1{/tex}or, {tex}n = 2m - 1{/tex}{tex}\\frac { a + ( m - 1 ) d } { A + ( m - 1 ) D } = \\frac { 7 ( 2 m - 1 ) + 1 } { 4 ( 2 m - 1 ) + 27 }{/tex}{tex}\\frac { a + ( m - 1 ) d } { A + ( m - 1 ) D } = \\frac { 14 m - 7 + 1 } { 8 m - 4 + 27 }{/tex}{tex}\\frac { a + ( m - 1 ) d } { A + ( m - 1 ) D } = \\frac { 14 m - 6 } { 8 m + 23 }{/tex}Hence,\xa0{tex}\\frac { a _ { m } } { A _ { m } } = \\frac { 14 m - 6 } { 8 m + 23 }{/tex}


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