Question

The number of ways to select three numbers in $A.P.$ from the first $2n$ natural numbers is

• A. $n(n-1)$
• B. ${}^{2n}{C}_n$
• C. $2n^2$
• D. $n^2$

Answer 1

Answer: (A)

$n(n-1)$

Let n=2,
Hence we get first four natural numbers.
Thus the A.P could be
$1,2,3$
$(2,3,4)$
Hence 2 ways
Hence $\left(2\right)(2-1)$ ways.
Let n=3,
we get first 6 natural numbers.
$1,2,3,4,5,6$
Hence the A.Ps will be
$(1,2,3)$
$(2,3,4)$
$(3,4,5)$
$(4,5,6)$
$(1,3,5)$
$(2,4,6)$
Hence there will be 6 ways of selecting an A.P
Hence
$\left(3\right)(3-1)$ ways.
Thus the number of ways of select 3 numbers in A.P out of first 2n numbers is $n(n-1)$