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if the ratio of sum of n terms of two APs is (3n+8):(7n+15), then the ratio of 12th terms is??
Sn = n/2(2a + (n-1)d)
(Sn)1/(Sn)2 = [2a1 + (n-1)d1] / [2a2 + (n-1)d2]
= [a1 + (n-1)d1/2 ] / [a2 + (n-1)d2/2]
this expression will become ratio of nth term if , (n-1)/2 is made (k-1) ...
(Sn)1/(Sn)2 = 3n+8/7n+15 (given)
for (n-1)/2 = k-1 or n = 2k-1
(Sn)1/(Sn)2 = (Tn)1/(Tn)2 = 3(2k-1) + 8/7(2k-1) + 15 = (6k + 5)/(14k + 8)
now put k=12 , this will give ration of 12th term
T12 = 77/176
approve my ans if u like it
thnx
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