Question

Determine the AP whose $3^{rd}$ term is 5 and the $7^{th}$ term is 9.

Answer 1

We have

$a_3$= a + (3 – 1) d = a + 2d = 5 (1)

and $a_7$ = a + (7 – 1) d = a + 6d = 9 (2)

Solving the pair of linear equations (1) and (2):
From equation(1)$-$ equation(2), we get, $a+2d-a-6d$ = 5-9
$-4d=-4\Rightarrow d=1$
Substitute $d=1$in equation(1), we get
$\Rightarrow a+2\left(1\right)=5$
$\Rightarrow a=5-2$
$\therefore a=3$

Thus, a = 3, d = 1

Hence, the required AP is 3, 4, 5, 6, 7….