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

The Correct Option is D

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:

\[ 3 + 2(4) = 11 \]