Question

If 4 times the 4th term of an arithmetic progression is equal to 9 times the 9th term of the same arithmetic progression, what is 13 times the 13th term?

• A. 0
• B. 30
• C. 60
• D. 120

Answer 1

The correct option is A 0

Let the first term of the arithmetic progression be $a$.
Let the common difference be $d$.

Let's use the term formula:
$t_n=a+(n-1)d$

The fourth term is:
$t_4=a+(4-1)d$
$t_4=a+3d$

The ninth term is:
$t_9=a+(9-1)d$
$t_9=a+8d$

$4t_4=9t_9$
$4(a+3d)=9(a+8d)$
$4a+12d=9a+72d$
$12d-72d=9a-4a$
$-60d=5a$
$-12d=a$

The thirteenth term is:
$t_{13}=a+(13-1)d$
$t_{13}=a+12d$
$t_{13}=-12d+12d$
$t_{13}=0$

The thirteenth term of the progression is 0.