if the sum of p terms of an A.p. is q and the sum of its q terms is p,then find the sum of its( p+q )term
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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)
hope it helps you!!!
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)
hope it helps you!!!
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