Question

Find the number of terms in the series $101 + 99 + 97 + ... + 47$ is
A) $25$
B) $28$
C) $30$
D) $20$

Answer 1

Given question is an arithmetic series. First we find the common distance and first term . After finding this we find the formula to find the $n$ th term. The formula of $n$ th term is ${a_n} = a + (n - 1)d$ , where $a$ is the first term and $d$ is the common difference between two consecutive terms . Common difference $=$ Second term $-$ first term. After putting the given value, calculate the value of $n$ .

Complete step by step answer:
First we find the first term of the arithmetic series
First term is $101$ and the second term is $99$ .
$\therefore$ Common difference is $99 - 101$
$= - 2$
Now we take the last term as the $n$ th term and we find the value of $n$ .
$\Rightarrow {a_n} = 47$
$\therefore$ The $n$ th term is ${a_n} = a + (n - 1)d$ , where $a$ is the first term and $d$ is the common difference between two consecutive terms
Put the value of $a = 101$ , ${a_n} = 47$ and $d = - 2$ in the above equation and get the value
$\Rightarrow 47 = 101 + (n - 1)( - 2)$
$\Rightarrow 47 = 101 - (n - 1) \times 2$
$\Rightarrow 2(n - 1) = 101 - 47$
We divide both sides of the above equation by $2$ and we get
$\Rightarrow n - 1 = 27$
$\Rightarrow n = 27 + 1$
$\Rightarrow n = 28$
Therefore the value of $n$ is $28$
$\therefore$ The number of terms in the series is $28$. Hence, option (B) is correct.

Note:
We can also solve the given problem by using the shortcut method . It is not a descriptive answer, it is a mcq answer . At first we find the common difference between two consecutive terms and after that we find the difference between the first and last term of the series . After that we find the $\dfrac{t}{s} + 1$, where $t$ is the difference between first and last term and $s$ is the common difference between two consecutive terms.