Question:
In the binomial expansion of $(x+a)^{15}$, if the eleventh term is the geometric mean of the eighth and twelfth terms, then the numerically greatest term in its expansion is
In the binomial expansion of $(x+a)^{15}$, if the eleventh term is the geometric mean of the eighth and twelfth terms, then the numerically greatest term in its expansion is
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For the greatest term, examine $\dfrac{T_{r+2}}{T_{r+1}}$ and locate where it crosses $1$.
Updated On: Jun 3, 2026
- $8$
- $9$
- $10$
- $11$
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The Correct Option is A
Solution and Explanation
Step 1: Concept
Use the condition for geometric mean and then determine the greatest term.
Step 2: Meaning
General term of \[ (x+a)^{15} \] is \[ T_{r+1}={}^{15}C_r x^{15-r}a^r. \] Given \[ T_{11}^2=T_8T_{12}. \]
Step 3: Analysis
Substituting, \[ \left({}^{15}C_{10}x^5a^{10}\right)^2 = \left({}^{15}C_7x^8a^7\right) \left({}^{15}C_{11}x^4a^{11}\right). \] Using \[ {}^{15}C_{10}={} ^{15}C_5, \qquad {}^{15}C_{11}={} ^{15}C_4, \] and simplifying, \[ a=x. \] Hence \[ \frac{a}{x}=1. \] For the greatest term, \[ \frac{T_{r+2}}{T_{r+1}} = \frac{15-r}{r+1}\cdot\frac{a}{x} = \frac{15-r}{r+1}. \] The greatest term occurs when \[ \frac{15-r}{r+1}\ge1. \] Thus \[ 15-r\ge r+1 \] \[ r\le7. \] Therefore the greatest term is \[ T_{8}. \]
Step 4: Conclusion
Hence the numerically greatest term is the $8^{\text{th}}$ term.
Final Answer: (A)
Use the condition for geometric mean and then determine the greatest term.
Step 2: Meaning
General term of \[ (x+a)^{15} \] is \[ T_{r+1}={}^{15}C_r x^{15-r}a^r. \] Given \[ T_{11}^2=T_8T_{12}. \]
Step 3: Analysis
Substituting, \[ \left({}^{15}C_{10}x^5a^{10}\right)^2 = \left({}^{15}C_7x^8a^7\right) \left({}^{15}C_{11}x^4a^{11}\right). \] Using \[ {}^{15}C_{10}={} ^{15}C_5, \qquad {}^{15}C_{11}={} ^{15}C_4, \] and simplifying, \[ a=x. \] Hence \[ \frac{a}{x}=1. \] For the greatest term, \[ \frac{T_{r+2}}{T_{r+1}} = \frac{15-r}{r+1}\cdot\frac{a}{x} = \frac{15-r}{r+1}. \] The greatest term occurs when \[ \frac{15-r}{r+1}\ge1. \] Thus \[ 15-r\ge r+1 \] \[ r\le7. \] Therefore the greatest term is \[ T_{8}. \]
Step 4: Conclusion
Hence the numerically greatest term is the $8^{\text{th}}$ term.
Final Answer: (A)
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