Question

Only question Number one

Answer 1

Let's start arbitrarily with the term a(1).
sum of first m terms = m/2 * (a(1) + a(m))
= m/2 * [a(1) + a(1) + (m-1)*d]
= m/2 * [2*a(1) + (m-1)*d] ...(1)

Sum of next n terms = n/2 * [a(m+1) + a(m+n)]
= n/2 * [a(1) + m*d + a(1) + (m+n-1)*d]
= n/2 * [2*a(1) + (2m+n-1)*d] ...(2)

Sum of next p terms = p/2 * [a(m+n+1) + a(m+n+p)]
= p/2 * [a(1) + (m+n)*d + a(1) + (m+n+p-1)*d]
= p/2 * [2*a(1) + (2m+2n+p-1)*d] ...(3)

All of these equal S. So now we need to eliminate S and a(1) and d.

S*(2/n - 2/m) = (m+n)*d ...(4)
S*(2/p - 2/n) = (n+p)*d ... (5)

Divide the two to get
(m+n)/(n+p) = (1/n - 1/m) / (1/p - 1/n)
(m+n) * (1/n -1/p) = (n+p) * (1/m - 1/n)


Hope this will help you.