If fourth term in the expansion of (px+1x)n is 52, find the value of 'n' and 'p'

T4=T3+1= nC3(px)n−3(1x)3⇒ nC3pn−3⋅xn−6=52x°⇒n−6=0⇒n=6⇒ nC3pn−3=52⇒ 6C3p3=52⇒p3=18∴p=12.

If fourth term in the expansion of (px+1x)n is 52, find the value of 'n' and 'p'

T4=T3+1= nC3(px)n−3(1x)3⇒ nC3pn−3⋅xn−6=52x°⇒n−6=0⇒n=6⇒ nC3pn−3=52⇒ 6C3p3=52⇒p3=18∴p=12.

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If fourth term in the expansion of ${(p x + \frac{1}{x})}^{n}$ is $\frac{5}{2}$ , find the value of $' n' and' p'$

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$n = 6, p = \frac{1}{2}$

$n = \frac{1}{2}, p = 6$

$n = 2, p = \frac{1}{6}$

$n = \frac{1}{2}, p = 2$

answer is A.

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Detailed Solution

$T_{4} = T_{3 + 1} =^{n} C_{3} {(p x)}^{n - 3} {(\frac{1}{x})}^{3}$

$\Rightarrow^{n} C_{3} p^{n - 3} \cdot x^{n - 6} = \frac{5}{2} x °$

$\Rightarrow n - 6 = 0 \Rightarrow n = 6$

$\Rightarrow^{n} C_{3} p^{n - 3} = \frac{5}{2}$

$\Rightarrow^{6} C_{3} p^{3} = \frac{5}{2}$

$\Rightarrow p^{3} = \frac{1}{8}$

$∴ p = \frac{1}{2}$ .

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If fourth term in the expansion of ${(p x + \frac{1}{x})}^{n}$ is $\frac{5}{2}$ , find the value of $' n' and' p'$