TS EAMCET · Maths · Binomial Theorem
Suppose \(l, m, n\) respectively represent the coefficient of \(x^{10}\), the constant term and the coefficient of \(x^{-10}\) in the expansion of \(\left(a x^2+\frac{b}{x^3}\right)^{15}\). If \(\frac{l}{m}+\frac{m}{n}=\frac{26}{11}\), then \(a^2: b^2=\)
- A \(16: 9\)
- B \(9: 4\)
- C \(4: 1\)
- D \(1: 25\)
Answer & Solution
Correct Answer
(B) \(9: 4\)
Step-by-step Solution
Detailed explanation
Given binomial \(\left(a x^2+\frac{b}{x^3}\right)^{15}\), so the general term, \(T_{r+1}={ }^{15} C_r a^{15-r} b^r x^{30-5 r}\) \(\therefore\) The coefficient of \(x^{10}\) is (at \(r=4\) ) \(={ }^{15} C_4 a^{11} b^4=l\) (given) Similarly,…
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