Question

Determine the AP whose third term is $16$ and 7th term exceeds the 5th term by $12$.

• A. $2,4,6,8,......$
• B. $3,9,15,21,......$
• C. $4,10,16,22,......$
• D. $5,10,15,20,......$

Answer 1

The correct option is C $4,10,16,22,......$

Let $a$ be the First term, $a_3$ be the third term, $a_5$ be the 5th term and $a_7$ be the 7th term

We know, $a_n=a+(n-1)d$

$\Rightarrow$ $a_3=16$ [ Given ]

$\Rightarrow$ $a+(3-1)d=16$

$\Rightarrow$ $a+2d=16$ ....... $\left(1\right)$

$\Rightarrow$ Now, $a_7-a_5=12$ [ Given ]

$\Rightarrow$ $[a+(7-1\left)d\right]-[a+(5-1\left)d\right]=12$

$\Rightarrow$ $2d=12$

$\therefore$ $d=6$

From equation $\left(1\right),$ we get

$a+2\left(6\right)=16$

$a+12=16$

$\therefore$ $a=4$

So first term is $4$

We can find AP by adding $d$ continuously.

$\therefore$ Required $AP$ is $4,10,16,22,28,34,40,......$