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n AMs are  inserted between 1 & 51. If ratio between 4th & 7th AMs is 3:5, then n equals??

8 years ago

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

8 years ago

thnx

8 years ago

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
one year ago
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