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CBSE Class 10 Answered

THE RATIO OF THE SUM OF n terms of two A.P. is(7n+1):(4n+27).find the ratio of mth terms
 
Asked by Shambhu Nath Tiwary | 15 Nov, 2015, 10:37: PM
answered-by-expert Expert Answer
Let space straight S subscript straight A space and space straight S subscript straight B space be space the space sums space straight n space terms space of space two space APs. Let space straight a subscript straight A space and space straight a subscript straight B space be space the space initial space terms space and space straight d subscript straight A space and space straight d subscript straight B space be space the space common space differences space of space two space APs. Given space that space the space ratio space of space the space sum space of space two space APs space is space open parentheses 7 straight n plus 1 close parentheses colon open parentheses 4 straight n plus 27 close parentheses rightwards double arrow straight S subscript straight A over straight S subscript straight B equals fraction numerator 7 straight n plus 1 over denominator 4 straight n plus 27 end fraction rightwards double arrow fraction numerator begin display style straight n over 2 open square brackets 2 straight a subscript straight A plus open parentheses straight n minus 1 close parentheses straight d subscript straight A close square brackets end style over denominator straight n over 2 open square brackets 2 straight a subscript straight B plus open parentheses straight n minus 1 close parentheses straight d subscript straight B close square brackets end fraction equals fraction numerator 7 straight n plus 1 over denominator 4 straight n plus 27 end fraction rightwards double arrow fraction numerator begin display style 2 straight a subscript straight A plus open parentheses straight n minus 1 close parentheses straight d subscript straight A end style over denominator 2 straight a subscript straight B plus open parentheses straight n minus 1 close parentheses straight d subscript straight B end fraction equals fraction numerator 7 straight n plus 1 over denominator 4 straight n plus 27 end fraction rightwards double arrow fraction numerator begin display style fraction numerator begin display style 2 straight a subscript straight A plus open parentheses straight n minus 1 close parentheses straight d subscript straight A end style over denominator 2 end fraction end style over denominator begin display style fraction numerator 2 straight a subscript straight B plus open parentheses straight n minus 1 close parentheses straight d subscript straight B over denominator 2 end fraction end style end fraction equals fraction numerator 7 straight n plus 1 over denominator 4 straight n plus 27 end fraction rightwards double arrow fraction numerator begin display style straight a subscript straight A plus fraction numerator begin display style open parentheses straight n minus 1 close parentheses straight d subscript straight A end style over denominator 2 end fraction end style over denominator begin display style straight a subscript straight B plus fraction numerator open parentheses straight n minus 1 close parentheses straight d subscript straight B over denominator 2 end fraction end style end fraction equals fraction numerator 7 straight n plus 1 over denominator 4 straight n plus 27 end fraction rightwards double arrow fraction numerator begin display style straight a subscript straight A plus fraction numerator begin display style open parentheses straight n minus 1 close parentheses end style over denominator 2 end fraction straight d subscript straight A end style over denominator begin display style straight a subscript straight B plus fraction numerator open parentheses straight n minus 1 close parentheses over denominator 2 end fraction straight d subscript straight B end style end fraction equals fraction numerator 7 straight n plus 1 over denominator 4 straight n plus 27 end fraction... left parenthesis 1 right parenthesis Assume space fraction numerator open parentheses straight n minus 1 close parentheses over denominator 2 end fraction equals open parentheses straight m minus 1 close parentheses rightwards double arrow straight n minus 1 equals 2 open parentheses straight m minus 1 close parentheses rightwards double arrow straight n minus 1 equals 2 straight m minus 2 rightwards double arrow straight n equals 2 straight m minus 2 plus 1 rightwards double arrow straight n equals 2 straight m minus 1 Thus comma space substituting space the space values comma space fraction numerator open parentheses straight n minus 1 close parentheses over denominator 2 end fraction equals open parentheses straight m minus 1 close parentheses space and space straight n equals 2 straight m minus 1 space in space equation space left parenthesis 1 right parenthesis comma space we space have comma fraction numerator begin display style straight a subscript straight A plus open parentheses straight m minus 1 close parentheses straight d subscript straight A end style over denominator begin display style straight a subscript straight B plus open parentheses straight m minus 1 close parentheses straight d subscript straight B end style end fraction equals fraction numerator 7 open parentheses 2 straight m minus 1 close parentheses plus 1 over denominator 4 open parentheses 2 straight m minus 1 close parentheses plus 27 end fraction rightwards double arrow straight t subscript mA over straight t subscript mB equals fraction numerator 14 straight m minus 7 plus 1 over denominator 8 straight m minus 4 plus 27 end fraction rightwards double arrow straight t subscript mA over straight t subscript mB equals fraction numerator 14 straight m minus 6 over denominator 8 straight m plus 23 end fraction Thus space the space ratio space of space straight m space terms space of space two space APs space is space 14 straight m minus 6 colon 8 straight m plus 23
Error converting from MathML to accessible text.
Hope this clears your doubt.
Answered by Vimala Ramamurthy | 14 Mar, 2017, 08:24: PM

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