Question

If first term of an A.P. is ‘p’, second term is ‘q’ and last term is ‘r’, then show that the sum of all terms is `((p + r)(q + r - 2p))/(2(q - p))`.

Answer 1

Proof:

Here, t₁ = p, t₂ = q and t_n = r.

The common difference (d) = t₂ – t₁ = q – p.

The total number of terms in the A.P.,

t_n = a + (n – 1)d   ...(Formula)

∴ r = p + (n – 1) × (q – p)

∴ r – p = (n – 1) × (q – p)

∴ (n – 1) × (q – p) = r – 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)`

The sum of n terms of the A.P.,

`S_n = n/2 (t_1 + t_n)`

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

= `((r + q - 2p)(p + r))/(2(q - p))`, i.e. `S_n = ((r + p)(q + r - 2p))/(2(q - p))`.