Question

If the sum of first p terms of an A.P. is equal to the sum of the first q terms, then
find the sum of the first (p + q) terms.

Answer 1

Answer:

0

Step-by-step explanation:

Sp = Sq (given)

∴ (p/2)[2a + (p-1)d] = (q/2)[2a + (q-1)d]

Multiplying both sides by 2, we get

p[2a + (p-1)d] = q[2a + (q-1)d]

2ap + (p-1)pd = 2aq + (q-1)qd

2ap - 2aq = (q-1)qd - (p-1)pd

2a(p-q) = q²d - qd - p²d + pd

2a(p-q) = q²d - p²d + pd  - qd

2a(p-q) = -d[p² - q²  - p + q]

2a(p-q) = -d[(p² - q²)  - (p - q)]

2a(p-q)= -d[(p+q)(p-q) - 1×(p -q)]

Taking out (p-q) as common from the Right Hand Side,(RHS)

2a(p-q) = -d(p-q)[p+q-1]

2a = -d(p+q-1)

2a + (p+q-1)d = 0 ........ eqn I

Now S(p+q) = {(p+q)/2} [2a + (p+q -1)d]

But 2a + (p+q-1)d = 0 from eqn I

∴ S(p+q) =  {(p+q)/2} [0] = 0

∴ sum of the first (p + q) terms is 0