Question

If sum of p terms of AP is q and sum of q terms of AP is p. Find the sum of (p+q) terms..

Answer 1

Dear mates here is your answer

➡️Let the first term of the given AP be ‘a' and the common difference be ‘d'. Then, the sum of first ‘n' terms of the AP is given by:

S(n)= n/2 {2a+(n-1)d} …….(1)

Here, it is given that:

S(p)=q and S(q)=p

Using (1), we get:-

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

and p= q/2 {2a+(q-1)d}

i.e. 2a+(p-1)d = 2q/p …..(2)

and 2a+(q-1)d = 2p/q …..(3)

Subtracting (3) from (2), we get:

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

So, d= 2(q^2-p^2)/pq(p-q)

i.e. d= -2(p+q)/pq

Now, substituting the value of ‘d' in eq.n (2), we get:

2a + (p-1){-2(p+q)/pq} = 2q/p

i.e. 2a= 2q/p + 2(p-1)(p+q)/pq

This gives:

a= (p^2+q^2-p-q+pq)/pq

So, we have

S(p+q)= (p+q)/2 { 2(p^2+q^2-p-q+pq)/pq - (p+q-1) 2 (p+q)/pq}

i.e. S(p+q)= (p+q)/pq { p^2+q^2-p-q+pq-p^2-pq-qp-q^2+p+q}

So, S(p+q) = -(p+q). … :-)

Answer 2

Let the first term of the given AP be ‘a' and the common difference be ‘d'. Then, the sum of first ‘n' terms of the AP is given by:

S(n)= n/2 {2a+(n-1)d} …….(1)

Here, it is given that:

S(p)=q and S(q)=p

Using (1), we get:-

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

and p= q/2 {2a+(q-1)d}

i.e. 2a+(p-1)d = 2q/p …..(2)

and 2a+(q-1)d = 2p/q …..(3)

Subtracting (3) from (2), we get:

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

So, d= 2(q^2-p^2)/pq(p-q)

i.e. d= -2(p+q)/pq

Now, substituting the value of ‘d' in eq.n (2), we get:

2a + (p-1){-2(p+q)/pq} = 2q/p

i.e. 2a= 2q/p + 2(p-1)(p+q)/pq

This gives:

a= (p^2+q^2-p-q+pq)/pq

So, we have

S(p+q)= (p+q)/2 { 2(p^2+q^2-p-q+pq)/pq - (p+q-1) 2 (p+q)/pq}

i.e. S(p+q)= (p+q)/pq { p^2+q^2-p-q+pq-p^2-pq-qp-q^2+p+q}

So, S(p+q) = -(p+q). … :-)