if the sum of the first P terms of an ap is Q and the sum of the first Q terms is P where p and q show that the sum of the first P + Q term is - p + q

Answer: Hence proved, Sp+q=−(p+q)S_{p+q} = -(p+q).

Step-by-step solution

Step 1: Set up equations for given sums

Let aa be the first term and dd be the common difference of the AP. Using the formula for the sum of the first nn terms, Sn=n2[2a+(n−1)d]S_n = \frac{n}{2}[2a + (n-1)d], we set up equations for Sp=qS_p = q and Sq=pS_q = p.

Step 2: Subtract the two equations to find d

Subtracting equation (2) from equation (1) eliminates 2a2a, giving (p−1)d−(q−1)d=(p−q)d(p-1)d - (q-1)d = (p-q)d. Factoring the right-hand side using q2−p2=−(p−q)(p+q)q^2 - p^2 = -(p-q)(p+q), we divide both sides by (p−q)(p-q) to solve for dd.

Step 3: Find the expression for 2a+(p+q−1)d2a + (p + q - 1)d

Alternatively, multiply equation (1) by pp to get 2ap+p(p−1)d=2q2ap + p(p-1)d = 2q and equation (2) by qq to get 2aq+q(q−1)d=2p2aq + q(q-1)d = 2p. Subtracting these gives 2a(p−q)+[p(p−1)−q(q−1)]d=2(q−p)2a(p-q) + [p(p-1) - q(q-1)]d = 2(q-p), which simplifies to 2a+(p+q−1)d=−22a + (p+q-1)d = -2 upon dividing throughout by (p−q)(p-q).

Step 4: Compute the sum of the first (p + q) terms

Substitute the bracket value 2a+(p+q−1)d=−22a + (p+q-1)d = -2 directly into the formula for Sp+qS_{p+q}. The factors of 22 cancel, leaving the final required sum −(p+q)-(p+q).

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