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Question

If n A. M.s are inserted between 20 and 80 such that first mean: last mean =1:3.Find the value of n

A
13
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B
12
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C
10
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D
11
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Solution

The correct option is D 11
a=20(first term) l=80(last\quad term)
no of AMs=n
total no of term in AP=no of means+one first term+one last term.
N=n+2
let the common difference =d
l=a+(N1)d80=20+(n+21)dd=60n+1.........(1)
Now, first mean=a+d=20+d
Last mean=a+nd=20+ndRatio=1:320+d20+nd=1360+3d=20+nd40=d(n3)d=40(n3)...............(2)
Equating (1)and(2),we get
60(n+1)=40(n3)60n180=40n+4020n=220=>n=11

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