Question

Positive integers $a_1,a_2,a_3,\dots \dots \dots$ form an A.P. If $a_1=10$ and $a_{a_2}=100$ then

• A. common difference of A.P. is equal to $6$.
• B. The value of a $a_3$ is equal to $20.$
• C. The value of $a_{22}$ is equal to $136$
• D. The value of $a_{a_3}$ is equal to $820$

Answer 1

The correct option is A common difference of A.P. is equal to $6$.

$a_1=10$

$a_{a_2}=100$

$100=a_1+(a_2-1)d$

$100=10+(a_2-1)d$

$90=(a_2-1)d$ (1)

$a_2=a_1+(2-1)d$

$a_2=10+d$ (2)

Substituting (2) in (1)

$90=(10+d-1)d$

$90=(9+d)d$

$90=9d+d^2$
$d^2+9d-90=0$

$d=\frac{-9\pm \sqrt{9^2+4\times 90}}{2}$

$d=\frac{-9\pm \sqrt{81+360}}{2}$

$d=\frac{-9\pm \sqrt{441}}{2}$

$d=\frac{-9\pm \sqrt{441}}{2}$

$d=\frac{-9\pm 21}{2}$

$d=6$ or $d=-15$

d cannot be negative.

Hence the correct option is (A).