Question

Ans the above ques plzz​

Answer 1

Answer:

Tp:Spq=1/2(pq+1)

Step-by-step explanation:

an=a+(n-1)d

pth=1/q

ap=1/q

a+(p-1)d=1/q ....(1)

aq=1/p

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

(1)-(2)

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

a+pq-d-[a+qd-d]=1/q-1/p

a+pd-d-a-qd+d=1/q-1/p

a-a-d+d+pd-qd=1/q-1/p

pd-qd=1/q-1/p

d(p-q)=1/q-1/p

d(p-q)=p-q/p-q

d=p-q/p-q(p-q)

d=1/p-q

Solving for a,

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

1/q=a+(p-1)1/p-q

1/q=a+p/pq-1/p-q

1/q=a+1/q-1/pq

1/q-1/q=a-1/pq=a-1/pq

a=1/pq

Spq=1/2(pq+1)

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

Spq=pq/2[2*1/pq+(pq-1)1/pq]

=pq/2[2/pq+pq/pq-1/pq]

=pq/2[1+pq/pq]

=1/2[pq+1]

Hence proved