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(T_n) = 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

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