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n AMs are inserted between 1 & 51. If ratio between 4th & 7th AMs is 3:5, then n equals??
first term of this AP series is 1 , so a =1 now nth term(Tn) = a+(n-1)d (d is common ration of this series) 4th term of AMs means 5th term of AP ...so T5 = a+4d = 1+ 4d 7th term of AMs means 8th term of AP ...so T8 = a+7d = 1+7d ratio is given = 3/5 1+4d/1+7d = 3/5 5+20d = 3+21d d = 2 now ,let there are n terms in this Ap then a + (n-1) = 51 1 + (n-1)2 = 51 or n = 26 approve my ans if u like it
first term of this AP series is 1 , so a =1
now nth term(Tn) = a+(n-1)d (d is common ration of this series)
4th term of AMs means 5th term of AP ...so T5 = a+4d = 1+ 4d
7th term of AMs means 8th term of AP ...so T8 = a+7d = 1+7d
ratio is given = 3/5
1+4d/1+7d = 3/5
5+20d = 3+21d
d = 2
now ,let there are n terms in this Ap then
a + (n-1) = 51
1 + (n-1)2 = 51 or
n = 26
approve my ans if u like it
thnx
There are n+2 terms in the AP because we’re also including 1 and 51 so the last step is wrong.It should be 1 + (n+1)d = 51 so n = 24
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