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Question

If Narithmetic means are inserted between 1and 31 such that ratio of first and nthmean is 3:29,then what is the value of N?


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Solution

Solve for value of N.

The first term is a=1 and the nth term is an=31

Number of A.M.'s =n

Total no. of terms in A.P =no. of means+ first term+ last term

N=n+2

Let common difference=d, an=a+(N-1)d

31=1+(n+2-1)d30=(n+1)d

d =30n+1...(1)

The first mean=1+d

The nthmean =1+nd

The ratio of first mean to the nth mean is 3:29,i.e.

1+d1+nd32929(1+d)=3(1+nd)29+29d=3+3nd26=3nd-29d26=(3n-29)dd=263n-29...(2)

By solving both the equations(1) and (2) we get:

30n+1=263n-2930(3n-29)=26(n+1)90n-870=26n+2690n-26n=26+87064n=896n=89664n=14

Hence, the value of Nis 14.


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