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CBSE Class 10 Answered

Page 255 Question 22 The ratio of the sum of n terms of two AP is (7n+1) : (4n+27). Find the ratio of their nth terms
Asked by miniprasad | 15 Dec, 2018, 07:15: PM
answered-by-expert Expert Answer
Let a1 and a2 be the first term  of two A.Ps .  Let d1 and d2 be the common differences of two A.Ps .
 
begin mathsize 12px style fraction numerator 2 a subscript 1 space plus space open parentheses n space minus space 1 close parentheses space d subscript 1 over denominator 2 a subscript 2 space plus space open parentheses n minus 1 close parentheses d subscript 2 end fraction space equals space fraction numerator 7 n plus 1 over denominator 4 n plus 27 end fraction l e t space u s space d i v i d e space t h e space d e n o m i n a t o r space a n d space n u m e r a t o r space o f space l e f t space h a n d space s i d e space b y space 2 fraction numerator begin display style a subscript 1 space plus space 1 half open parentheses n space minus space 1 close parentheses space d subscript 1 end style over denominator begin display style a subscript 2 space plus 1 half open parentheses n minus 1 close parentheses d subscript 2 end style end fraction space equals space fraction numerator begin display style 7 n plus 1 end style over denominator begin display style 4 n plus 27 end style end fraction space space l e t space u s space s u b s t i t u t e space 1 half left parenthesis n minus 1 right parenthesis space equals space m minus 1 space i n space t h e space L H S space o f space a b o v e space e q n. fraction numerator begin display style a subscript 1 space plus space open parentheses m space minus space 1 close parentheses space d subscript 1 end style over denominator begin display style a subscript 2 space plus space open parentheses m minus 1 close parentheses d subscript 2 end style end fraction space equals space fraction numerator begin display style 7 n plus 1 end style over denominator begin display style 4 n plus 27 end style end fraction L H S space o f space a b o v e space e q u a t i o n space i s space t h e space r a t i o space o f space m to the power of t h end exponent space t e r m space o f space t w o space A. P s. f r o m space s u b s t i t u t i o n comma space w e space h a v e space n space equals space 2 m minus 1. space U sin g space t h i s space r e l a t i o n comma w e space c a n space e x p r e s s space t h e space r i g h t space h a n d space s i d e space o f space a b o v e space e q u a t i o n space i n space t e r m s space o f space m space space H e n c e space space fraction numerator begin display style a subscript 1 space plus space open parentheses m space minus space 1 close parentheses space d subscript 1 end style over denominator begin display style a subscript 2 space plus space open parentheses m minus 1 close parentheses d subscript 2 end style end fraction space equals space fraction numerator begin display style 7 m space minus space 3 end style over denominator begin display style 4 m space minus space 1 end style end fraction o r space r a t i o space o f space n t h space t e r m space o f space t w o space A. P s space space equals space fraction numerator begin display style 7 n space minus space 3 end style over denominator begin display style 4 n minus space 1 end style end fraction end style
Answered by Thiyagarajan K | 17 Dec, 2018, 10:12: AM

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