Question

Show that the sum of (p+q)th and (p-q)th terms of an A.P is equal to twice the pth term

Answer 1

Sol:

Let a be the first term and d is common difference of the A.P then sum of n terms in A.P

is Sn = (n/2)[ 2a + (n - 1) d]

Given that Sp = q and Sq = p.

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

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

Subtract (1) from (2) we get

(q - p)d = (2p / q) – (2q / p)

(q - p)d = (2p2– 2q2) / pq

d = -2(q +p) / pq -----------(3)

Sum of first ( p + q ) terms

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

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

Sp +q = (p +q) / 2 [ (2q /p) + q(-2(q +p) / pq )] [ from (1) and (3)]

Sp +q = (p + q) / 2 [ (2q -2q-2p) / p )]

Sp +q = (p + q) / 2 [-2p) / p )]

Sp +q = - (p + q).

Answer 2

Answer:

Step-by-step explanation:

Please see the attached answer