Question

If the 3rd , 4th , 5th and 6th terms in the expansion of
$\huge \bf{ \red{ \mid{ \overline{ \underline{(x + \alpha )^{n} }}} \mid}}$
be respectively a , b , c and d , prove that

$\pink {\huge{ {\bf{ \frac{ {b}^{2} - ac }{ {c}^{2} - bd } = \frac{5a}{3c}}}}}$
​

Answer 1

Hey!

Given :

T3 = a,

T4 = b,

T5 = c,

T6 = d.

Refer attachment :

-> Applying basic concepts of Binomial Theorem!

=)

$5 \div 3 \: = \: (b ^{2} - ac) \times \: c \: \div \: c ^{2} - bd \: \times \: a$

Hence b² - ac / c² - bd = 5a/3c.

Answer 2

5/3 = (b² - ac) × c/c² - bd × a

b² - ac / c² - bd

= 5a / 3c.