Question

The number of terms in the series $101+99+97+...+47$ is

• A. $25$
• B. $28$
• C. $30$
• D. $20$

Answer 1

The correct option is B

$28$

Calculate the number of terms in the given arithmetic progression

The given series is $101+99+97+...+47$.

Let, there are $n$ terms in the given series.

Here, the first term $\left(a\right)=101$

The second term $\left(a_2\right)=99$

The last term $\left(a_n\right)=47$

And, the common difference $\left(d\right)=99-101$ $\left[\because d=a_2-a_1\right]$

$\Rightarrow$ $d=-2$

As we know, the $n^{th}$ of an AP (Arithmetic Progression) is given by,

$a_n=a+\left(n-1\right)d$

So, by substituting the values of $a_n,a$ and $d$ in the above, we get,

$47=101+\left(n-1\right)\left(-2\right)$

$\Rightarrow$ $47=101-2n+2$

$\Rightarrow$ $2n=56$

$\Rightarrow$ $n=28$

that is, there are $n=28$ terms in the given series.

Hence, option (B) is the correct option.