SATHEE: Binomial Theorem 1 Question 36

Binomial Theorem 1 Question 36

38. The coefficients of three consecutive terms of $(1 + x)^{n + 5}$ are in the ratio $5 : 10 : 14$ . Then, $n$ is equal to (2013 Adv.)

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Solution:

Let the three consecutive terms in $(1 + x)^{n + 5}$ be $t_{r}, t_{r + 1}, t_{r + 2}$ having coefficients

$^{n + 5} C_{r - 1},^{n + 5} C_{r},^{n + 5} C_{r + 1}$ .

Given, $^{n + 5} C_{r - 1} :^{n + 5} C_{r} :^{n + 5} C_{r + 1} = 5 : 10 : 14$

$∴ \frac{^{n + 5} C_{r}}{^{n + 5} C_{r - 1}} = \frac{10}{5}$ and $\frac{^{n + 5} C_{r + 1}}{^{n + 5} C_{r}} = \frac{14}{10}$

$\Rightarrow \frac{n + 5 - (r - 1)}{r} = 2$ and $\frac{n - r + 5}{r + 1} = \frac{7}{5}$

$\Rightarrow n - r + 6 = 2 r$ and $5 n - 5 r + 25 = 7 r + 7$

$\Rightarrow n + 6 = 3 r$ and $5 n + 18 = 12 r$

$∴ \frac{n + 6}{3} = \frac{5 n + 18}{12}$

$\Rightarrow 4 n + 24 = 5 n + 18 \Rightarrow n = 6$

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Binomial Theorem 1 Question 36

38. The coefficients of three consecutive terms of (1+x)n+5 are in the ratio 5:10:14. Then, n is equal to (2013 Adv.)

Solution:

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38. The coefficients of three consecutive terms of $(1 + x)^{n + 5}$ are in the ratio $5 : 10 : 14$ . Then, $n$ is equal to (2013 Adv.)