If the pth term of an A.P. is q and the qth term is p, prove that its nth term is (p + q – n).
Given: ap = q and aq = p
To Prove: an = p + q - n
We know that,
an = a + (n – 1)d
Where, a = first term, d = common difference and n = number of terms.
Therefore,
pth term = ap
ap = a + (p – 1)d = q…..(1)
qth term = aq
aq = a + (q – 1)d = p….(2)
Subtracting equation 2 from equation 1, we get,
(p – 1)d – (q – 1)d = q – p
pd – d – qd + d = q – p
(p – q)d = (q – p)
d = - 1
Putting the value of ‘d’ in equation (1), we get
a + (p – 1)(-1) = q
a – p + 1 = q
a = p + q – 1
Now, putting the values of ‘a’ and ‘d’ in the formula for nth term, we get,
an = (p + q – 1) + (n – 1)(-1)
an = p + q – 1 – n + 1
an = p + q – n
Hence, Proved.
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