If the sum of m terms of an A.P. is equal to half the sum of (m+n) ter

1.	If the sum of m terms of an A.P. is equal to half the sum of (m+n) terms and isalso equal to half of the sum of (m+p) terms.Prove that: (m+n) (-) = (m+p) (-).
Answer» 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 [2a(1) + (m-1)d] ...(1) Sum of next n terms = n/2 * [a(m+1) + a(m+n)]= n/2 * [a(1) + md + a(1) + (m+n-1)d]= n/2 * [2a(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 * [2a(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)

If the sum of m terms of an A.P. is equal to half the sum of (m+n) terms and isalso equal to half of the sum of (m+p) terms.Prove that: (m+n) (-) = (m+p) (-).

Answer»

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)