Question

show that the sum of (P+q) and (P-q)th term of an AP is equal to twice its p th 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:2tp=2tp

Step-by-step explanation:tp+q+tp-q=2tp

As we know that :tn=a+(n-1)d so

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

Similarly

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

Acoding to condition

Tp+q+Tp-q=2Tp

Putting the values of tp+q and tp-q it will becomes

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

a+dp+dq-d+a+dp-dq-d

+dq-dq cancel

2a+2dp-2d=2Tp

Taking 2 common

2(a+dp-d)=2Tp

Again d is common

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

Tp=a+(p-1)d

So

2Tp=2Tp proved!!!