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:

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

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

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