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))`.

Answer 1

Given: First term (t₁) = a = p, Second term (t₂) = q, t_n = r

Common difference (d) = t₂ – t₁ = q – p

According to the question,

t_n = 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

S_n = `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]`

S_n = `(q + r - 2p) xx ((p + r))/(2(q - p))`

Hence proved.