First slide

Binomial theorem for positive integral Index

Question

Ratio of middle terms in ${(p x^{3} + \frac{q}{x^{2}})}^{15}$ is

Easy

A

$p : q x^{2}$
B

$p x^{5} : q$
C

$p x^{2} : q$
D

$p : q x^{5}$

Solution

Given expansion ${(p x^{3} + \frac{q}{x^{2}})}^{15}$ is
Index of the expansion $n = 15$ then middle terms are $\frac{15 + 1}{2} = 8, \frac{15 + 3}{2} = 9$
$8^{t h}$ term:
$T_{r + 1} =^{n} C_{r} x^{n - r} {(a)}^{r}$
$T_{8} = T_{7 + 1} = 15_{c_{7}} {(p x^{3})}^{8} {(\frac{q}{x^{2}})}^{7}$
$T_{8} = T_{7 + 1} = 15_{c_{7}} {(p)}^{8} {(x^{3})}^{8} {(q)}^{7} {(x^{- 2})}^{7}$
$9^{t h}$ term:
$T_{r + 1} =^{n} C_{r} x^{n - r} {(a)}^{r}$
$T_{9} = T_{8 + 1} = 15_{c_{8}} {(p x^{3})}^{7} {(\frac{q}{x^{2}})}^{8}$
$T_{9} = T_{8 + 1} = 15_{c_{8}} {(p)}^{7} {(x^{3})}^{7} {(q)}^{8} {(x^{- 2})}^{8}$
Ratio of middle terms in ${(p x^{3} + \frac{q}{x^{2}})}^{15}$ is
$\frac{T_{8}}{T_{9}} = \frac{T_{7 + 1}}{T_{8 + 1}} = \frac{15_{c_{7}} {(p)}^{8} {(x^{3})}^{8} {(q)}^{7} {(x^{- 2})}^{7}}{15_{c_{8}} {(p)}^{7} {(x^{3})}^{7} {(q)}^{8} {(x^{- 2})}^{8}}$
$= \frac{p x^{3}}{\frac{q}{x^{2}}} = p x^{5} : q$

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