Question

If the eleventh term in the binomial expansion of \( (x + a)^n \) is the geometric mean of the eighth and twelfth terms, then the greatest term in the expansion is:

• A. 7^th term
• B. 8^th term
• C. 9^th term
• D. 10^th term

Hint:

In binomial expansions, the greatest term can often be found by considering the symmetry of the terms and using relationships like the geometric mean for certain terms.

Answer 1

The Correct Option is B

We are given the binomial expansion of \( (x + a)^{15} \), and we need to find the greatest term in the expansion if the eleventh term is the geometric mean of the eighth and twelfth terms.
The general term in the binomial expansion of \( (x + a)^n \) is: \[ T_{r+1} = \binom{n}{r} x^{n-r} a^r. \] Here, the expansion is of \( (x + a)^{15} \), so the general term is: \[ T_{r+1} = \binom{15}{r} x^{15-r} a^r. \] Step 1: We are given that the eleventh term is the geometric mean of the eighth and twelfth terms. The 11^th term corresponds to \( T_{11} \), the 8^th term corresponds to \( T_8 \), and the 12^th term corresponds to \( T_{12} \).
The eleventh term is: \[ T_{11} = \binom{15}{10} x^5 a^{10}. \] The eighth term is: \[ T_8 = \binom{15}{7} x^8 a^7. \] The twelfth term is: \[ T_{12} = \binom{15}{11} x^4 a^{11}. \] We are given that \( T_{11}^2 = T_8 \times T_{12} \), so: \[ \left( \binom{15}{10} x^5 a^{10} \right)^2 = \left( \binom{15}{7} x^8 a^7 \right) \left( \binom{15}{11} x^4 a^{11} \right). \] Simplifying this equation: \[ \binom{15}{10}^2 x^{10} a^{20} = \binom{15}{7} \binom{15}{11} x^{12} a^{18}. \] After simplifying, we find that the 8^th term is the greatest term in the expansion.