Let (x+10)^ 5 +(x-10)^ 5 a =a_ a +a_ 1 x+a_ 2 x^ 2 + +a_ 5 a x^ 5 a , for all x R; then…
Mathematics · JEE Main · NTA Exams — Binomial Theorem And Its Simple Applications
Let \((x+10)^{5}+(x-10)^{5 a}=a_{a}+a_{1} x+a_{2} x^{2}+\ldots+a_{5 a} x^{5 a}\), for all x ∈ R; then \(\frac{a_{2}}{a_{n}}\) is equal to:
- 12.25
- 12.75
- 12.00
- 12.50
Answer
(D) 12.50
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