Q16 of 202 Page 10

In an AP, S_p = q, S_q = p and S_r denotes the sum of first r terms. Then, S_{p + q} is equal to

S_p = {2A + (p–1) D} = q, given

S_q = (2A + (q–1) D} = p, given

On Subtracting the second equation from 1st we get,

D = –2

Also On adding the two equations we get,

2A + = p/q + q/p

Now,

S_{p + q} = {2A + (p + q–1) D}

= {2A + (p + q–1)D}

{ + – + (p + q–1) D}

[ By substituting the value of D ]

= – (p + q)

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