Question

If $a,b,c$ and $d$ are any four consecutive coefficients in the expansion of $(1+x{)}^n$, 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. A.G.P.

Answer 1

The correct option is C H.P.

Given expansion is $(1+x{)}^n$
Let $a,b,c$ and $d$ be the $(r+1{)}^{th},(r+2{)}^{th},(r+3{)}^{th}$ and $(r+4{)}^{th}$ coefficients respectively, so
$a={}^{n}Cr,b={}^n{C}_{r+1}c={}^n{C}_{r+2},d={}^n{C}_{r+3}$

Now,
$\frac{a+b}{a}=\frac{{}^n{C}_r+{}^n{C}_{r+1}}{{}^n{C}_r}\Rightarrow \frac{a+b}{a}=\frac{{}^{n+1}{C}_{r+1}}{{}^n{C}_r}=\frac{n+1}{r+1}\frac{b+c}{b}=\frac{{}^n{C}_{r+2}+{}^n{C}_{r+1}}{{}^n{C}_{r+1}}\Rightarrow \frac{b+c}{b}=\frac{{}^{n+1}{C}_{r+2}}{{}^n{C}_{r+1}}=\frac{n+1}{r+2}$
Similarly,
$\frac{c+d}{c}=\frac{n+1}{r+3}$
Now, by observation, ​​​​​
$\frac{a}{a+b}+\frac{c}{c+d}=\frac{2r+4}{n+1}\Rightarrow \frac{a}{a+b}+\frac{c}{c+d}=2\times \frac{b}{b+c}$

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