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CBSE - X - Mathematics - Arithmetic Progression

About an A.P sum.............

Asked by 25th February 2008, 3:09 PM
Answered by Expert




Your question was 
The ratio of the sum of n terms of two AP is (4n+2): (3n+47).find the ratio of the 9th term.

Let the I term and common difference of the AP be a and d and for the II AP it will be 
a' and d' respectively 
According to the given Condition 
 n/2(2a+(n-1)d):  n/2(2a'+(n-1)d') = (4n+2): (3n+47)
i.e (2a+(n-1)d):  (2a'+(n-1)d') =(4n+2): (3n+47)
i.e (a+(n-1)d/2):(a'+(n-1)d/2)=(4n+2): (3n+47)
substituting n=17 in both sides we get




(a+8d):(a'+8d') = 70:98

(a+8d):(a'+8d') = 5:7

Answered by Expert 5th December 2017, 5:43 PM

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