Question

Find the sum of series: $ 1 + 2 + 3 + 4 + 5 + ..... + 100 $ .

Answer 1

Hint: The given problem requires us to find the sum of an arithmetic progression. The first few terms and the last term of the series is given to us in the question. For finding out the sum of an arithmetic progression, we need to know the first term, the common difference and the number of terms in the arithmetic progression. We can find out the common difference of an arithmetic progression by knowing the difference of any two consecutive terms of the series.

Complete step-by-step answer:
So, we have, $ 1 + 2 + 3 + 4 + 5 + ..... + 100 $
The difference of any two consecutive terms of the given series is constant. So, the given sequence is an arithmetic progression.
Now, we have to find the sum of this arithmetic progression.
Here, first term $ = a = 1 $ .
Now, we can find the common difference of the arithmetic progression by subtracting any two consecutive terms in the series.
So, common ratio \[ = d = 2 - 1 = 1\]
So, $ d = 1 $ .
We also need to know the number of terms in the arithmetic progression in order to find the value of the sum of arithmetic progression.
Also, we are given that the last term of the AP is $ 100 $ .
We know that the formula for the nth terms of an AP is $ {a_n} = a + \left( {n - 1} \right)d $ .
So, considering $ 100 $ as the nth term of the AP, we can find the number of terms in the AP and then evaluate the sum of the arithmetic progression.
So, $ {a_n} = a + \left( {n - 1} \right)d = 100 $
Substituting the values of a and d in the above equation, we get,
 $ \Rightarrow 1 + \left( {n - 1} \right)1 = 100 $
Now, solving the above equation for the value of n, we get,
 $ \Rightarrow n = 100 $
So, there are $ 100 $ terms in the given arithmetic progression.
Now, we can find the sum of the given arithmetic progression using the formula $ S = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right] $
Hence, the sum of AP $ = S = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right] $
Substituting the values of a, d and n in the formula, we get,
 $ \Rightarrow S = \dfrac{{100}}{2}\left[ {2\left( 1 \right) + \left( {100 - 1} \right)\left( 1 \right)} \right] $
Opening the brackets and cancelling the common factors in numerator and denominator, we get,
 $ \Rightarrow S = 50\left[ {2 + 99\left( 1 \right)} \right] $
Simplifying the expression, we get,
 $ \Rightarrow S = 50\left[ {101} \right] $
 $ \Rightarrow S = 5050 $
So, the sum of the given series: $ 1 + 2 + 3 + 4 + 5 + ..... + 100 $ is $ S = 5050 $ .
So, the correct answer is “5050”.

Note: Arithmetic progression is a series where any two consecutive terms have the same difference between them. The common difference of an arithmetic series can be calculated by subtraction of any two consecutive terms of the series. The sum of n terms of an arithmetic progression can be calculated if we know the first term, the number of terms and difference of the arithmetic series as: $ S = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right] $ .