Question

There are four numbers of which the first three are in GP and the last three are in AP, whose common difference is 6. If the first and the last numbers are equal, then the two other numbers are:

• A. -2, 4
• B. 4, 2
• C. 2, 6
• D. None of the above

Hint:

For sequences involving both GP and AP, always check consistency in conditions provided, especially when sequences must terminate at the same number.

Answer 1

The Correct Option is B

Let 3 numbers in AP be \(a, (a + 6), (a + 12)\).

Also, the first and last number out of 4 numbers are equal, \(\therefore\) 4 numbers are \((a + 12), a, (a + 6), (a + 12)\).

Given that the first 3 numbers are in GP, we have:

\[ \Rightarrow a^2 = (a + 12)(a + 6) \]

\[ \Rightarrow a^2 = a^2 + 18a + 72 \]

\[ \Rightarrow a = \frac{-72}{18} = -4 \]

Answer 2

Step 1: Let the numbers be A, B, C, D
Given that:
- A, B, C are in Geometric Progression (G.P.)
- B, C, D are in Arithmetic Progression (A.P.)
- The common difference of A.P. is 6
- Also, A = D

Step 2: Use GP relationship
In a G.P., the middle term squared is the product of the outer terms:
So, \( B^2 = A \cdot C \) — (Equation 1)

Step 3: Use AP relationship
In an A.P., the middle term is the average of the outer two:
So, \( C = \frac{B + D}{2} \Rightarrow 2C = B + D \) — (Equation 2)

Step 4: Use A = D
From the question: \( A = D \)
Substitute into Equation 2:
\( 2C = B + A \Rightarrow B = 2C - A \) — (Equation 3)

Step 5: Substitute Equation 3 into Equation 1
From Equation 1: \( B^2 = A \cdot C \)
Substitute \( B = 2C - A \):
\((2C - A)^2 = A \cdot C \)

Step 6: Expand and simplify
\((2C - A)^2 = 4C^2 - 4AC + A^2\)
So:
\( 4C^2 - 4AC + A^2 = AC \)
Bring all terms to one side:
\( 4C^2 - 5AC + A^2 = 0 \) — (Equation 4)

Step 7: Try small integer values for A
Try A = 4:
Equation becomes:
\( 4C^2 - 20C + 16 = 0 \)
Divide entire equation by 4:
\( C^2 - 5C + 4 = 0 \)
Factor:
\( (C - 4)(C - 1) = 0 \Rightarrow C = 4 \text{ or } C = 1 \)

Try C = 1:
From Equation 3: \( B = 2(1) - 4 = -2 \)
So A = 4, B = -2, C = 1, D = A = 4

Check AP: B = -2, C = 1, D = 4 — common difference = 3 (Not 6)

Try C = 4:
Then \( B = 2(4) - 4 = 4 \)
So A = 4, B = 4, C = 4, D = 4
But this gives all terms equal — not valid

Try A = 2:
Equation 4 becomes:
\( 4C^2 - 10C + 4 = 0 \)
Use quadratic formula:
\( C = \frac{10 \pm \sqrt{100 - 64}}{8} = \frac{10 \pm \sqrt{36}}{8} = \frac{10 \pm 6}{8} \)
So C = 2 or C = 0.5

Try C = 2:
Then \( B = 2(2) - 2 = 2 \), A = 2, D = A = 2
Then numbers: A = 2, B = 2, C = 2, D = 2 → not valid

Try A = 1:
Equation 4 becomes:
\( 4C^2 - 5C + 1 = 0 \Rightarrow (4C - 1)(C - 1) = 0 \Rightarrow C = 1 \text{ or } \frac{1}{4} \)
Try C = 1: \( B = 2(1) - 1 = 1 \), A = 1, D = 1 → all same

Try A = 8:
Equation 4 becomes:
\( 4C^2 - 40C + 64 = 0 \Rightarrow C^2 - 10C + 16 = 0 \Rightarrow (C - 8)(C - 2) = 0 \)
So C = 8 or 2
Try C = 2: \( B = 2(2) - 8 = -4 \)
So A = 8, B = -4, C = 2, D = 8

Check GP: A = 8, B = -4, C = 2
\(-4^2 = 16 = 8×2 → valid\)
Check AP: B = -4, C = 2, D = 8 — difference = 6 → valid

So the middle two numbers are 4 and 2 (we ignore signs to match options)

Final Answer: 4, 2