# If the p<su

Given: ap = q and aq = p

To Prove: an = (p + q – n)

Proof:

We know that,

an = a + (n – 1)d

Putting n = p

ap = a + (p – 1)d

q = a + (p – 1)d

q – (p – 1)d = a …(i)

Similarly,

aq = a + (q – 1)d

p = a + (q – 1)d

p – (q – 1)d = a …(ii)

From eq. (i) and (ii), we get

q – (p – 1)d = p – (q – 1)d

q – pd + d = p – qd + d

qd – pd = p – q

d(q – p) = -(q – p)

d = -1

Putting the value of d in eq. (i), we get

q – (p – 1)(-1) = a

q + p – 1 = a

Now, we have to find nth term

We know that,

an = a + (n – 1)d

Putting a = (q + p – 1) and d = -1

an = (p + q – 1) + (n – 1)(-1)

an = p + q – 1 – n + 1

an = p + q – n

Hence Proved

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