Question

If $p\ne q$ and the sequences p, a, b, q and p, m, n, q each are in AP, then $\frac{b-a}{n-m}$ is

• A. $\frac{2}{3}$
• B. $\frac{3}{2}$
• C. 1
• D. $\frac{3}{4}$

Answer 1

The correct option is C

1

Given, p, a, b, q are in AP.

Using Arithmetic mean , we have
$\therefore$ 2a = p + b
$\Rightarrow$ 2a - b = p . . . (i)
and 2b = a + q
$\Rightarrow$ 2b - a = q . . . .(ii)

Also given p, m, n, q are in AP
$\therefore$ 2m = p + n
$\Rightarrow$ 2m - n = p . . . (iii)
and 2n = m + q
$\Rightarrow$ 2n - m = q . . . (iv)

From eq (i) & (iii), we get

2a - b = 2m - n . . . (v)

From eq (ii) & (iv), we get

2b - a = 2n - m . . . (vi)

Subtracting (vi) from (v), we get

2a - b - (2b - a) = 2m - n - (2n - m)
2a - b - 2b + a = 2m - n - 2n + m
3a - 3b = 3m - 3n
a - b = m - n
b - a = n - m
$\frac{b-a}{n-m}=1$