Question

If the number of ways of selecting $3$ numbers out of $1,2,3,\dots ,2n+1$ such that they form an increasing arithmetic progression is $441,$ then the sum of the divisors of $n$ is equal to

• A. $21$
• B. $32$
• C. $45$
• D. $60$

Answer 1

The correct option is B $32$

$a,b,c$ are in A.P.
$\Rightarrow a+c=2b$
$\Rightarrow$ Sum of two numbers is even.
$\Rightarrow$ Both numbers are either even or odd.

Odd numbers are $1,3,5,\cdots ,2n+1[\text{Total}n+1\text{numbers}]$
Even numbers are $2,4,6,\cdots ,2n\left[\text{Total}n\text{numbers}\right]$

$\therefore$ Required number of ways $={}^{n+1}{C}_2+{}^n{C}_2$
$\therefore {}^{n+1}{C}_2+{}^n{C}_2=441$
$\Rightarrow n^2=441$
$\Rightarrow n=21=3\times 7$

$\therefore$ Sum of divisors $=1+3+7+21=32$