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Question

The sums of 'n' terms of two arithmetic progression are in the ration 5n+4:9n+6 , find the ration of their 18th terms.

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Solution

Let the two A.P s be
a,(a+d),(a+2d).............(1)
Where, a is first term and d is common difference for first A.P
and A(A+D).(A+2D), ....... (2)
Where, A is first term and D is common difference for second A.P
Given that ,
sum of the first n terms of(1)Sum of first n term of (2)=5n+49n+6
n2[2a+(n1)d]n2[2A+(n1)D]=5n+49n+6
[2a+(n1)d][2A+(n1)D]=5n+49n+6
[a+(n1)2d][A+(n1)2D]=5n+49n+6......(i)
Since, the ratio of 18th terms is a18A18=a+17dA+17D....(ii)
Now [a+(n1)2d][A+(n1)2D] compare with a+17dA+17D, we get
(n1)2=17
n1=34
n=34+1
n=35
Substitute n=35 in equation(i), we get
[a+(351)2d][A+(351)2D]=5(35)+49(35)+6
a+17dA+17D=175+4315+6
a18A18=179321 [From equation(ii)]
Hence, the ratio of 18th term of first A.P and 18th term second A.P is 179:321.


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