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The ratio between the sum of two APs is (3m-18) : (7n-15) find the ratio of their 12th term |
| Answer» For the first AP,Let first term=a1common difference=d1using formula:{tex}\\Longrightarrow \\mathrm { S } _ { \\mathrm { n } } = \\frac { \\mathrm { n } } { 2 } [ 2 \\mathrm { a } + ( \\mathrm { n } - 1 ) \\mathrm { d } ]{/tex}{tex}\\Longrightarrow \\left( \\mathrm { S } _ { \\mathrm { n } } \\right) _ { 1 } = \\frac { \\mathrm { n } } { 2 } \\left[ 2 \\mathrm { a } _ { 1 } + ( \\mathrm { n } - 1 ) \\mathrm { d } _ { 1 } \\right]{/tex}For 2nd AP.Given,\xa0{tex}\\Rightarrow{/tex}No. of terms=nLet,\xa0{tex}\\Rightarrow{/tex}first term=a2{tex}\\Rightarrow{/tex}common difference=d2Using formula:{tex}\\Longrightarrow \\mathrm { S } _ { \\mathrm { n } } = \\frac { \\mathrm { n } } { 2 } [ 2 \\mathrm { a } + ( \\mathrm { n } - 1 ) \\mathrm { d } ]{/tex}{tex}\\Longrightarrow \\left( \\mathrm { S } _ { \\mathrm { n } } \\right) _ { 2 } = \\frac { \\mathrm { n } } { 2 } \\left[ 2 \\mathrm { a } _ { 2 } + ( \\mathrm { n } - 1 ) \\mathrm { d } _ { 2 } \\right]{/tex}According to question :{tex}\\Longrightarrow \\frac { \\left( \\mathrm { S } _ { \\mathrm { n } } \\right) _ { 1 } } { \\left( \\mathrm { S } _ { \\mathrm { n } } \\right) _ { 2 } } = \\frac { 3 \\mathrm { n } + 8 } { 7 \\mathrm { n } + 15 }{/tex}{tex}\\Rightarrow{/tex}{tex}\\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 { 3 n + 8 } { 7 n + 15 }{/tex}Substitute n=23:{tex}\\Longrightarrow \\frac { 2 a _ { 1 } + ( 23 - 1 ) d _ { 1 } } { 2 a _ { 2 } + ( 23 - 1 ) d _ { 2 } } = \\frac { 3 \\times 23 + 8 } { 7 \\times 23 + 15 }{/tex}{tex}\\Longrightarrow \\frac { 2 \\mathrm { a } _ { 1 } + 22 \\mathrm { d } _ { 1 } } { 2 \\mathrm { a } _ { 2 } + 22 \\mathrm { d } _ { 2 } } = \\frac { 69 + 8 } { 161 + 15 }{/tex}{tex}\\Longrightarrow \\frac { 2 \\left( a _ { 1 } + 11 d _ { 1 } \\right) } { 2 \\left( a _ { 2 } + 11 d _ { 2 } \\right) } = \\frac { 77 } { 176 }{/tex}{tex}\\Longrightarrow \\frac { a _ { 1 } + ( 12 - 1 ) d _ { 1 } } { a _ { 2 } + ( 12 - 1 ) d _ { 2 } } = \\frac { 7 } { 16 }{/tex}{tex}\\Longrightarrow \\frac { \\left( T _ { 12 } \\right) _ { 1 } } { \\left( T _ { 12 } \\right) _ { 2 } } = \\frac { 7 } { 16 }{/tex}{tex}\\therefore{/tex}\xa0(T12)1 : (T12)2=7: 16. | |