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.
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2
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
Answered by
1
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
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