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Question
If the first term of an A.P. is p, second term is q and last term is r, then show that sum of all terms is `(q + r - 2p) xx ((p + r))/(2(q - p))`.
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Solution
Given: First term (t1) = a = p, Second term (t2) = q, tn = r
Common difference (d) = t2 – t1 = q – p
According to the question,
tn = a + (n – 1) × d
r = p + (n – 1) × (q – p)
(r – p) = (n – 1) × (q – p)
n – 1 = `(r - p)/(q - p)`
n = `(r - p)/(q - p) + 1`
n = `(r - p + q - p)/(q - p)`
n = `(r + q - 2p)/(q - p)`
We know
Sn = `n/2[2a + (n - 1)d]`
= `(r + q - 2p)/(2(q - p)) [2p + ((r + q - 2p)/(q - p) - 1) (q - p)]`
= `(r + q - 2p)/(2(q - p)) [2p + ((r + q - 2p - (q - p))/(q - p)) (q - p)]`
= `(r + q - 2p)/(2(q - p)) [((r + q - 2p - q + p)/(q - p)) (q - p)]`
= `(r + q - 2p)/(2(q - p))[2p + ((r - p)/(q - p)) (q - p)]`
= `(r + q - 2p)/(2(q - p))[2p + r - p]`
= `(r + q - 2p)/(2(q - p))[r + p]`
Sn = `(q + r - 2p) xx ((p + r))/(2(q - p))`
Hence proved.
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