Question

Let 3, a, b, c be in A.P. and 3, a – 1, b + l, c + 9 be in G.P. Then, the arithmetic mean of a, b and c is :

• A. -4
• B. -1
• C. 13
• D. 11

Answer 1

Answer: (D)

11

Answer 2

Given that 3, $a, b, c$ are in arithmetic progression (A.P.), we know that the common difference is constant:

$a - 3 = b - a = c - b$

Thus, we have:

$a = 3 + d, \quad b = 3 + 2d, \quad c = 3 + 3d$

Next, we are given that 3, $a - 1, b + 1, c + 9$ are in geometric progression (G.P.), so the ratios of consecutive terms are equal:

$\frac{a - 1}{3} = \frac{b + 1}{a - 1} = \frac{c + 9}{b + 1}$

Let the common ratio be $r$, so:

$\frac{a - 1}{3} = r \quad \text{and} \quad \frac{b + 1}{a - 1} = r$

This gives:

$a - 1 = 3r, \quad b + 1 = (a - 1)r$

Substitute $a = 3 + d$:

$(3 + d) - 1 = 3r \implies d + 2 = 3r \implies r = \frac{d + 2}{3}$

Now, solve for $b$ and $c$:

$b = 3 + 2d, \quad c = 3 + 3d$

Finally, the arithmetic mean of $a, b, c$ is:

$\frac{a + b + c}{3} = \frac{(3 + d) + (3 + 2d) + (3 + 3d)}{3} = \frac{9 + 6d}{3} = 3 + 2d$

Given that $d = 4$, the arithmetic mean is:

$3 + 2(4) = 11$