Question

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

• A. Arithmetic Progression
• B. Geometric Progression
• C. Harmonic Progression
• D. None of these

Answer 1

The correct option is C

Harmonic Progression

Explanation for the correct option:

Step 1: Calculate the values of $\frac{a+b}{a},\frac{b+c}{b},\frac{c+d}{c}$.

We know $(x+y{)}^n={}^n{C}_0{x}^n+{}^n{C}_1{x}^{n-1}y+{}^n{C}_2{x}^{n-2}{y}^2+\dots ..{}^n{C}_n{y}^n$ where $a={}^n{C}_0,b={}^n{C}_1,c={}^n{C}_2,d={}^n{C}_3$

$\begin{array}{rlrlrl}a& =1& & & & \\ b& =n& & & & \\ c& =\frac{n(n-1)}{2}& & & & \\ d& =\frac{n(n-1)(n-2)}{6}& & & & \\ \frac{a+b}{a}& =1+n& & & & \\ \frac{b+c}{b}& =\frac{n+\frac{\left(n\right(n-1)}{2}}{n}& & & & \\ & =1+\frac{n-1}{2}& & & & \\ & =\frac{1+n}{2}& & & & \\ \frac{c+d}{c}& =\frac{\frac{n(n-1)}{2}+\frac{n(n-1)(n-2)}{6}}{\frac{n(n-1)}{2}}& & & & \\ & =1+\frac{n-2}{3}& & & & \\ & =\frac{1+n}{3}& & & & \end{array}$

Step 2: Check if the terms are in Arithmetic Progression, Geometric Progression or Harmonic Progression

Reciprocal of $\frac{a+b}{a},\frac{b+c}{b}\text{and}\frac{c+d}{c}$ are $\frac{1}{1+n},\frac{2}{1+n}\text{and}\frac{3}{1+n}$ respectively.

Since $\frac{2}{1+n}-\frac{1}{1+n}=\frac{1}{1+n}$ and $\frac{3}{1+n}-\frac{2}{1+n}=\frac{1}{1+n}$ so the difference is equal.

Also,

$\begin{array}{rcl}\frac{2\left[{\displaystyle \frac{(a+b)}{a}}\right]\left[{\displaystyle \frac{(c+d)}{c}}\right]}{\left[\frac{(a+b)}{a}\right]+\left[\frac{(c+d)}{c}\right]}& =& \frac{{\displaystyle \frac{2(1+n)(1+n)}{3}}}{{\displaystyle \frac{(1+n)+(1+n)}{3}}}\\ & =& \frac{{\displaystyle \frac{2(1+n)(1+n)}{3}}}{{\displaystyle \frac{(3+3n+1+n)}{3}}}\\ & =& \frac{{\displaystyle \frac{2(1+n)(1+n)}{3}}}{{\displaystyle \frac{(4+4n)}{3}}}\\ & =& \frac{2(1+n)(1+n)}{4(1+n)}\\ & =& \frac{(1+n)}{2}\\ & =& \frac{(b+c)}{b}\end{array}$

So $\frac{a+b}{a},\frac{b+c}{b},\frac{c+d}{c}$ are in Harmonic Progression.

Hence, Option ‘C’ is Correct.