Question

If the coefficients of three consecutive terms in the expansion of (1 + x)²³ are in arithmetic progression, then those terms are ?

• A. \(T_{10},\,T_{11},\,T_{12}\)
• B. \(T_{8},\,T_{9},\,T_{10}\)
• C. \(T_{13},\,T_{14},\,T_{15}\)
• D. \(T_{14},\,T_{15},\,T_{16}\)

Hint:

For \(\binom{n}{r}\) to form an arithmetic progression with consecutive \(r\)-values, leverage Pascal’s identity or test likely middle indices.
- Checking a few “middle” terms often solves such a binomial-coefficient A.P. problem quickly.

Answer 1

The Correct Option is D

Step 1: General binomial term.
In \((1 + x)^{n}\), the \((r+1)\)th term is \[ T_{r+1} = \binom{n}{r} x^r. \] For \((1 + x)^{23}\), the \(r\)th term can be written as: \[ T_r = \binom{23}{r-1}x^{r-1}, \quad \text{where } r=1,2,\dots,24. \] The coefficient of \(T_r\) is \(\binom{23}{r-1}\). Step 2: Arithmetic progression condition.
Suppose the three consecutive terms are \(T_k,\, T_{k+1},\,T_{k+2}\). Their coefficients must satisfy: \[ 2\,\binom{23}{k} = \binom{23}{k-1} + \binom{23}{k+1}. \] Use the identity \(\binom{n}{r-1} + \binom{n}{r} = \binom{n}{r}\times \frac{n-r+1}{r} + \binom{n}{r}\) carefully or Pascal’s rule to simplify. By standard checking, one finds \(k=14\) satisfies this. Step 3: Conclusion.
Hence the three consecutive terms with coefficients in A.P. are \(T_{14}, T_{15}, T_{16}\).