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If the ratio of the sum of the first n terms of two A.P.s is (7n+1) . . .

Question : If the ratio of the sum of the first n terms of two A.P.s is (7n+1):(4n+27), then find the ratio of their 9th terms.


Doubt by Shaurya

Solution : 

Let us consider two AP's having first term and common difference as a and a' and d and d' respectively. 

Sn:Sn'=(7n+1):(4n+27)
n/2[2a+(n-1)d] : 
n/2[2a'+(n-1)d']=(7n+1):(4n+27)
[2a+(n-1)d]:[2a'+(n-1)d']=(7n+1):(4n+27)
2[a+{(n-1)/2}d]:2[a'+{(n-1)/2}d']=(7n+1):(4n+27)
[a+{(n-1)/2}d]:[a'+{(n-1)/2}d']=(7n+1):(4n+27) — (1) 

Now in order to find the ratio of 9th term
{(n-1)/2} should be equal to 8
{(n-1)/2}=8
n-1=16
n=16+1
n=17

Putting n=17 in equation (1) 

a+8d:a'+8d=[7(17)+1]:[4(17)+27]
a+(9-1)d : a'+(9-1)d'=[119+1] : [68+27]
a9:a'9=120:95
a9:a'9=24:19

Hence, the ratio of 9th term is 24:19


Similar Question
If the ratio of the sum of first n terms of two AP's is (7n+1):(4n+27), find the ratio of their mth terms.