Question

If $a,b,c$ and $d$ are any four consecutive coefficients of any binomial expansion, then $\frac{a+b}{a},\frac{b+c}{b},\frac{c+d}{c}$ are in

• A. $A.P.$
• B. $G.P.$
• C. $H.P.$
• D. None of these

Answer 1

The correct option is C $H.P.$

Let the expansion be that of ${(1+x)}^n.$

Let $a,b,c,d$ be the $(r+1)$th, $(r+2)$th, $(r+3)$th and $(r+4)$th coefficients.

$\therefore a{=}^n{C}_r,b{=}^n{C}_{r+1},c{=}^n{C_{r+2},d=}^n{C}_{r+3.}$

Now, $\frac{a}{a+b}=\frac{{}^n{C}_r}{{}^n{C_r+}^n{C}_{r+1}}=\frac{{}^n{C}_r}{{}^{n+1}{C}_{r+1}}=\frac{n!}{r!(n-r)!}\times \frac{(r+1)!(n-r)!}{(n+1)!}=\frac{r+1}{n+1}$

Similarly, $\frac{b}{b+c}=\frac{(r+1)+1}{n+1}=\frac{r+2}{n+1},$

$\frac{c}{c+d}=\frac{(r+2)+1}{n+1}=\frac{r+3}{n+1}.$

$\therefore \frac{a}{a+b}+\frac{c}{c+d}=\frac{r+1}{n+1}+\frac{r+3}{n+1}=\frac{2r+4}{n+1}=\frac{2(r+2)}{n+1}=\frac{2b}{b+c}.$

$\Rightarrow \frac{a}{a+b},\frac{b}{b+c},\frac{c}{c+d}$ are in $A.P.$

$\therefore \frac{a+b}{a},\frac{b+c}{b},\frac{c+d}{c}$ are in $H.P.$