Question

The sum and product of three consecutive terms of an AP is given to be 21 and 168. If the common difference is known to be postitive, then the ${10}^{\text{th}}$ term of the AP is ___.

• A. 52
• B. 38
• C. 45
• D. 58

Answer 1

The correct option is A 52

Let the terms of the AP be $a-d,a$ and $a+d.$
By question, we have
$a-d+a+a+d=21.$
$\Rightarrow 3a=21$
$\Rightarrow a=7$
Also, by question, we have $(a-d)a(a+d)=168.$
i.e., $(7-d)7(7+d)=168$
$\Rightarrow 7(49-d^2)=168$
$\Rightarrow 343-7d^2=168$
$\Rightarrow 7d^2=175$
$\Rightarrow d^2=25$
$\Rightarrow d=\pm 5$
$d>0\Rightarrow d=5$
The $n^{\text{th}}$ term of an AP of first term $a$ and common difference $d$ is given by
$a_n=a+(n-1)d.$
$\Rightarrow a_{10}=a+9d=7+9\left(5\right)=52$