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:

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

Answer 2

Given that 3, a, b, c are in A.P. and 3, a–1, b+1, c+9 are in G.P., find the arithmetic mean of a, b, c.

Concept Used:

For an A.P. with first term 3 and common difference d: a = 3+d, b = 3+2d, c = 3+3d.
For a G.P. with first term 3 and common ratio r: a–1 = 3r, b+1 = 3r², c+9 = 3r³.

Step-by-Step Solution:

Step 1: Express a, b, c in terms of d using A.P.

Let common difference = d.
Then: a = 3 + d, b = 3 + 2d, c = 3 + 3d.

Step 2: Use G.P. condition: 3, a–1, b+1, c+9 are in G.P.

Let common ratio = r.
Then: a – 1 = 3r ...(1)
b + 1 = 3r² ...(2)
c + 9 = 3r³ ...(3)

Step 3: Substitute a, b, c from Step 1 into (1), (2), (3).

From (1): (3 + d) – 1 = 3r ⇒ 2 + d = 3r ...(1')
From (2): (3 + 2d) + 1 = 3r² ⇒ 4 + 2d = 3r² ...(2')
From (3): (3 + 3d) + 9 = 3r³ ⇒ 12 + 3d = 3r³ ⇒ 4 + d = r³ ...(3')

Step 4: Solve for r and d.

From (1'): r = (2 + d)/3.
From (2'): 3r² = 4 + 2d ⇒ r² = (4 + 2d)/3.
But r² = [(2 + d)/3]² from (1').
So: [(2 + d)/3]² = (4 + 2d)/3
Multiply by 9: (2 + d)² = 3(4 + 2d)
4 + 4d + d² = 12 + 6d
d² – 2d – 8 = 0
(d – 4)(d + 2) = 0 ⇒ d = 4 or d = –2.

Step 5: Check both cases using (3').

For d = 4: r = (2+4)/3 = 2, r³ = 8.
From (3'): 4 + d = 4 + 4 = 8 = r³. ✓ Works.
For d = –2: r = (2–2)/3 = 0, r³ = 0.
From (3'): 4 + (–2) = 2 ≠ 0. ✗ Fails.
So d = 4, r = 2.

Step 6: Find a, b, c and their arithmetic mean.

a = 3 + 4 = 7, b = 3 + 8 = 11, c = 3 + 12 = 15.
Arithmetic mean = (a + b + c)/3 = (7 + 11 + 15)/3 = 33/3 = 11.

Therefore, the arithmetic mean of a, b, c is 11.