Question

Statement 1: The $n^{th}$ term of an AP can not be $n^2+1$.
Statement 2: 0 is a term of the AP 31, 28, 25….

• A. Both statement 1 and
  statement 2 are true.
• B. Statement 1 is true, while statement 2 is false.
• C. Statement 1 is false, while statement 2 is true.
• D. Both statement 1 and statement 2 are false.

Answer 1

The correct option is B Statement 1 is true, while statement 2 is false.

Statement 1: The $n^{th}$ term of an AP cannot be $n^2$ + 1.

$n^{th}term=n^2+1$.
$1^{st}term=1\times 1+1$
= 2
$2^{nd}term=2\times 2+1$
= 4+1 =5
$3^{rd}term=3\times 3+1$
= 10
common difference = $2^{nd}$ term - $1^{st}$ term = 3 $\ne$ $3^{rd}$ term - $2^{nd}$ term = 5.

That means Statement 1 is true.

Statement 2: 0 is a term of the AP 31, 28, 25….
$n^{th}$ term = a + (n - 1)d.
$1^{st}$ term (a) = 31
$2^{nd}$ term = 28
common difference (d) = $2^{nd}$ term - $1^{st}$ term
= 28 - 31 = -3
Let’s assume that nth term of this AP is 0.
$n^{th}$ term = a + (n - 1)d = 0
= 31 - (n - 1)3 = 0
$\Rightarrow n=\frac{34}{3}$
$n$ should be a natural number. In this case n is a fraction that means 0 is not a term in given AP.