Question

If $a,b,c,d$ are any four consecutive coefficients in the expansion of $(1+x{)}^n$, then $\frac{a}{a+b},\frac{b}{b+c},\frac{c}{c+d}$ are in:

• A. A.P.
• B. G.P.
• C. H.P.
• D. A.G.P

Answer 1

The correct option is A A.P.

$a={}^n{C}_r$
$b={}^n{C}_{r+1}$
$c={}^n{C}_{r+2}$
$d={}^n{C}_{r+3}$
Therefore
$\frac{a}{a+b}$ $=\frac{{}^n{C}_r}{{}^n{C}_r+{}^n{C}_{r+1}}$ $=\frac{{}^n{C}_r}{{}^{n+1}{C}_r}$

Upon simplifying, we get
$\frac{{}^n{C}_r}{{}^{n+1}{C}_r}=1-\frac{r}{n+1}$ ...(i)
$\frac{b}{b+c}$ $=\frac{{}^n{C}_{r+1}}{{}^n{C}_{r+1}+{}^n{C}_{r+2}}$
$=\frac{{}^n{C}_{r+1}}{{}^{n+1}{C}_{r+1}}$
Upon simplifying, we get
$\frac{{}^n{C}_{r+1}}{{}^{n+1}{C}_{r+1}}=1-\frac{r+1}{n+1}$ ...(ii)
$\frac{c}{c+d}$ $=\frac{{}^n{C}_{r+2}}{{}^n{C}_{r+2}+{}^n{C}_{r+3}}$
$=\frac{{}^n{C}_{r+2}}{{}^{n+1}{C}_{r+2}}$
Upon simplifying, we get
$\frac{{}^n{C}_{r+2}}{{}^{n+1}{C}_{r+2}}=1-\frac{r+2}{n+1}$ ...(iii)

From (i), (ii) and (iii) it can be shown that $\frac{a}{a+c},\frac{b}{b+c},\frac{c}{c+d}$ are in A.P.