Question

If the average of the first '$n$' numbers in the sequence $148,146,144,...$ is $125$, then $n=$

• A. $18$
• B. $24$
• C. $30$
• D. $36$

Answer 1

Answer: (B)

$24$

The sequence $148,146,\mathrm{144......}$ is an A.P

$\therefore$ sum of $n$ terms = $\frac{n}{2}(2a+(n-1\left)d\right)$

Now $a=148$

$d=-2$

$\therefore s=\frac{n}{2}(2\times 148+(n-1)-2)$

$\therefore s=\frac{n}{2}(296-2n+2)$

$\therefore \frac{s}{n}=\frac{1}{2}(296-2n+2)$

$\therefore \frac{s}{n}=149-n$

Now average = $\frac{s}{n}=125\to$ Given

$\therefore 149-n=125$

$\therefore n=149-125$

$\therefore n=24$

$\therefore$ Number of terms is $24$