if the pth term of AP is q and its qth term is p then show that its (p+q)th is zero.......
Answers
EXPLANATION.
- GIVEN
P th term of an Ap = q
q th term of an Ap = P
Show that it's ( P + q) th term os zero.
according to the question,
Pth term = q
=> a + ( P - 1 ) d = q ......(1)
qth term = P
=> a + ( q - 1 ) d = P ......(2)
From equation (1) and (2) we get,
we get,
=> ( P - q) d = q - P
=> d = -1
put the value of d = -1 in equation (1)
we get,
=> a + ( P - 1 ) -1 = q
=> a + ( -P + 1 ) = q
=> a - P + 1 = q
=> a = q + P - 1
( P + q ) th term
=> a + ( P + q - 1 ) d
=> q + P - 1 + ( P + q - 1 ) (-1)
=> q + P - 1 + ( - P - q + 1 )
=> q + P - 1 - P - q + 1
=> 0
Therefore,
( P + q ) th term is zero
HENCE PROVED.
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