Question

If the 4th and 6th terms of an arithmetic progression are 21 and 33, respectively, find the 13th term.

• A. 72
• B. 83
• C. 85
• D. 75

Answer 1

The correct option is D 75

Terms in an arithmetic progression:

$T_n=a+(n–1)d$
Here, a = first term and d = common difference.

$T_4=a+(4–1)d$

$T_4=a+3d$ ------(1)

$T_6=a+(6–1)d$

$T_6=a+5d$ ------(2)

Solving (1) and (2), we get:

d = 6 and a = 3

$T_{13}=a+(13–1)d$

Substituting the values of 'a' and 'd', we get:

$T_{13}=3+\left(12\right)6$

$T_{13}=3+72$

$T_{13}=75$