Greatest term of this expansion?

Question

If in the expansion of $(x+a)^{15}$ the $11^{\text{th}}$ term is G. M. i.e. geometric mean of $8^{\text{th}}$ and $12^{\text{th}}$ terms then which term is the greatest term among the terms in expansion of $(x+a)^{15}$?

Answer 1

Applying the binomial theorem we have \begin{align*} (x+a)^{15}=\sum_{j=0}^{15}\binom{15}{j}x^ja^{15-j} \end{align*}

Since the $11^{\text{th}}$ term is the geometric mean of the the $8^{\text{th}}$ and $12^{\text{th}}$ term, we have \begin{align*} \binom{15}{10}x^{10}a^5&=\sqrt{\binom{15}{7}x^7a^8\binom{15}{11}x^{11}a^4}\\ &=x^9a^6\sqrt{\binom{15}{7}\binom{15}{11}}\\ \end{align*}

Simplification with $a\ne 0$ gives \begin{align*} \frac{x}{a}&=\frac{\sqrt{\binom{15}{7}\binom{15}{11}}}{\binom{15}{10}}=\ldots=\frac{5}{77}\sqrt{231}\approx0.98\tag{1} \end{align*}

In order to find the greatest term we compare successive terms we obtain \begin{align*} \binom{15}{j+1}x^{j+1}a^{14-j}&>\binom{15}{j}x^ja^{15-j}\tag{2}\\ \end{align*} Simplification of (2) gives \begin{align*} \frac{x}{a}&>\binom{15}{j}\binom{15}{j+1}^{-1}=\ldots=\frac{j+1}{15-j}\\ \end{align*}

After rearrangement we get $j$ on the left-hand side and evaluation according (1) gives \begin{align*} j&<\frac{15\frac{x}{a}-1}{\frac{x}{a}+1}\bigg|_{\frac{x}{a}=\frac{5}{77}\sqrt{231}}\approx 6.8\tag{3} \end{align*}

We conclude $j=6$ and the greatest term is according to (2) the term at the left-hand side of (2) which is (with $j+1=7$) the $8^{\text{th}}$ term \begin{align*} \color{blue}{\binom{15}{7}x^7a^8} \end{align*}