Question

The pth term of an ap is q and the qth term of this ap is p ,show that its (p+q)th term is zero.

Answer 1

Answer:

ap= a+(p-1)d =q and aq= a+(q-1)d =p

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

= 2a+pd-d+ qd-d =q+p

= 2a+ d(p+q -2) -(q+p) =0

= 2a + p+q{(d-2)-1} =0

= 2a + p+q{(d-3} =0

=(p+q)th =0

Answer 2

According to the Question

Let a be first term be a

And Common Difference be d

Therefore

\bf\huge a_{p} = q , a_{q} = pa

p

=q,a

q

=p

a + (p - 1)d = q ……. (1)

a + (q - 1)d = p ……..(2)

Subtracting equations we get :-

(p - q)d = q - p

d = -1

Put the value of d in eq (1) :-

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

a = (p + q - 1)

\bf\huge a_{p + q} = a + (p + q - 1)da

p+q

=a+(p+q−1)d

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

= 0

Hence we get the (p + q)th term is Zero