Solve this The third and ninth term of an arithmetic progression (AP) are 9 and 27 respectively find the fifth term

The formula for the $n$-th term of an arithmetic progression is $a_n = a_1 + (n-1)d$, where $a_1$ is the first term and $d$ is the common difference.

For the third term ($a_3 = 9$):

$$a_3 = a_1 + (3-1)d$$ $$9 = a_1 + 2d \quad (*)$$

For the ninth term ($a_9 = 27$):

$$a_9 = a_1 + (9-1)d$$ $$27 = a_1 + 8d \quad (**)$$

Subtract equation $(*)$ from equation $(**)$:

$$(a_1 + 8d) - (a_1 + 2d) = 27 - 9$$ $$6d = 18$$ $$d = \frac{18}{6}$$ $$d = 3$$

Substitute $d=3$ into equation $(*)$:

$$9 = a_1 + 2(3)$$ $$9 = a_1 + 6$$ $$a_1 = 9 - 6$$ $$a_1 = 3$$

Using the formula $a_n = a_1 + (n-1)d$ with $n=5$, $a_1=3$, and $d=3$:

$$a_5 = a_1 + (5-1)d$$ $$a_5 = 3 + 4(3)$$ $$a_5 = 3 + 12$$ $$a_5 = 15$$

The fifth term is $\text{15}$.