If a,b,c be the sum of n terms of three AP's whose first terms are unity and common difference are in H.P. then n=
a.(2ac+ab+bc)/(a+c-2b)
b. (2ac-ab-bc)/(a+c-2b)
c. (2ac-ab-bc)/(a+c+2b)
d. (2ac-ab+bc)/(a+c+2b)

Dear Vaibhav

let common difference are d1,d2 ,d3

a = n/2[2 +(n-1)d1]  or    1/d1 = n(n-1)/2(a-n)

similerly     1/d2 = n(n-1)/2(b-n)

                1/d3 = n(n-1)/2(c-n)

d1 ,d2 ,d3 are in HP

2/d2 = 1/d1 +1/d2

 n(n-1)/(b-n) = n(n-1)/2(a-n)   +  n(n-1)/2(c-n)

   1/(b-n) = 1/2(a-n)   +  1/2(c-n)

n= (2ac -ab-bc)/(a+c-2b)

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