Question

The P^th,q^th and r^th terms of an AP are a,b,c respectively. Show that (a-b)r + (b-c)P + (c-a)q = 0.​

Answer 1

Solution:

Let us assume that the first term of the given AP be A and the common difference be d.

Therefore, we have :-

$\rm \longrightarrow A_{p} = a$

$\rm \longrightarrow A_{q} = b$

$\rm \longrightarrow A_{r} = c$

Now, we know that :-

$\rm \longrightarrow A_{n} = a + (n - 1)d$

Therefore, we have :-

$\rm \longrightarrow a = A + (p - 1)d - (i)$

$\rm \longrightarrow b = A + (q - 1)d - (ii)$

$\rm \longrightarrow c = A + (r - 1)d - (iii)$

Subtracting (ii) from (i), we get :-

$\rm \longrightarrow (a - b) = (p - 1)d - (q - 1)d$

$\rm \longrightarrow (a - b) = (p - 1 - q + 1)d$

$\rm \longrightarrow (a - b) = (p- q)d$

$\rm \longrightarrow (a - b)r = rd(p- q) - (iv)$

Subtracting (iii) from (ii), we get :-

$\rm \longrightarrow (b - c) = (q - 1)d - (r- 1)d$

$\rm \longrightarrow (b - c) = (q - 1 - r + 1)d$

$\rm \longrightarrow (b - c) = (q- r)d$

$\rm \longrightarrow (b - c)p = pd(q- r) - (v)$

Subtracting (i) from (iii), we get :-

$\rm \longrightarrow (c - a) = (r - 1)d - (p - 1)d$

$\rm \longrightarrow (c - a) = (r - 1 - p + 1)d$

$\rm \longrightarrow (c - a) = (r- p)d$

$\rm \longrightarrow (c - a)q = qr(r- p) - (vi)$

Adding equations (iv), (v) and (vi), we get :-

$\rm \longrightarrow (a - b)r + (b - c)p + (c - a)q = rd(p - q) + pd(q - r) + qd(r - p)$

$\rm \longrightarrow (a - b)r + (b - c)p + (c - a)q =d \{r(p - q) + p(q - r) + q(r - p) \}$

$\rm \longrightarrow (a - b)r + (b - c)p + (c - a)q =d \{pr - qr + pq - pr + qr - pq \}$

$\rm \longrightarrow (a - b)r + (b - c)p + (c - a)q =d \{ - qr + pq + qr - pq \}$

$\rm \longrightarrow (a - b)r + (b - c)p + (c - a)q =d \{ + pq - pq \}$

$\rm \longrightarrow (a - b)r + (b - c)p + (c - a)q =d \times 0$

$\rm \longrightarrow (a - b)r + (b - c)p + (c - a)q =0$

Hence Proved..!!

Answer 2

Answer:

Refer to the attachments for working.