Question

In a certain progression three consecutive terms are $30,24,20$ . The next term of the progression is

• A. $16$
• B. $\frac{120}{7}$
• C. $18$
• D. $Noneofthese$

Answer 1

Answer: (B)

Given terms is $30,24,20$, here we observe that,
$\frac{1}{30},\frac{1}{24},\frac{1}{20}$ are in $AP$
The common difference
$d=\frac{1}{24}-\frac{1}{30}=\frac{1}{120}$
and first term $a=\frac{1}{30}$
Then, the fourth term is $T_4=a+3d$
$=\frac{1}{30}+3.\frac{1}{120}$
$=\frac{1}{30}+\frac{1}{40}=\frac{7}{120}$
Hence, the next term of progression $30,24,20$ is $\left(\frac{120}{7}\right)$