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`        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)`
7 years ago

147 Points
```
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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7 years ago
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