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. $None\, of\, these$

Answer 1

Answer: (B)

$\frac{120}{7}$

Answer 2

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 $\left(\frac{120}{7}\right)$