Question

In a sequence of 21 terms, the first 11 terms are in AP with common difference 2 and the last 11 terms are in GP with common ratio 2. If the middle term of the AP is equal to the middle term of the GP, then the middle term of the entire sequence is:

• A. \( -\frac{10}{31} \)
• B. \( \frac{10}{31} \)
• C. \( \frac{32}{31} \)
• D. \( -\frac{31}{32} \)

Hint:

In problems involving sequences of terms in AP and GP, carefully use the middle terms of each sequence and equate them if required. Use the general formulas for AP and GP to express the terms and solve for unknowns.

Answer 1

The Correct Option is A

Since the first 11 terms are an arithmetic progression (AP) with common difference \(d = 2\): \[ a_{11} = a + 10d = a + 20 \] The middle term of AP is: \[ T_6 = a + 5d = a + 10 \] For the next 11 terms in a geometric progression (GP) with common ratio \(r = 2\): \[ {The middle term of GP is } b(2^5) { where } b { is the first term of GP which is the last term of AP} \] \[ b(2^5) = (a + 20) \cdot 32 \] According to the given condition: \[ a + 10 = (a + 20) \cdot 32 \] \[ \Rightarrow 32a = 10 - 640 \] \[ \Rightarrow a = -\frac{630}{31} \] Thus, the middle term of the entire sequence is the 11th term: \[ T_{11} = -\frac{630}{31} + 10 \times d = -\frac{630}{31} + 10 \times 2 = -\frac{10}{31} \] Therefore, the correct answer is Option A.