Question

Find the sum: $25 + 28 + 31 + ..... + 100$

Answer 1

Hint: When we have to find the sum of any series which have equal difference between their consecutive terms than the sum is carried out by using the formula ${S_n} = \dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right)$ where, n is the number of terms in series, a and d are for first term and common difference.

Complete step-by-step answer:
Here, the given series is: $25 + 28 + 31 + ..... + 100$
The first term of the series is: 25
The common difference of the series is carried by using the formula given below:
${\rm{common}}\;{\rm{difference = second}}\;{\rm{term - first}}\;{\rm{term}}$
So, substituting the values in the formula then we get,
$d = 28 - 25 = 3$
Here, d stands for common difference.
Now, the total number of terms in the series is not given so we check that the last term comes on which place in the series which will show us the total number of terms of the given series.
Here, the last term of the series is: 100
The formula to find the last term of the series is: $l = a + \left( {n - 1} \right)d$
Where, $l$ is the last term of the series
Substituting the values in $l = a + \left( {n - 1} \right)d$then we get,
$
 \Rightarrow 100 = 25 + \left( {n - 1} \right)3\\
 \Rightarrow 100 = 25 + 3n - 3\\
 \Rightarrow 100 = 22 + 3n\\
 \Rightarrow 100 - 22 = 3n
$
Further calculating the expression then,
$
 \Rightarrow 78 = 3n\\
 \Rightarrow 3n = 78\\
 \Rightarrow n = \dfrac{{78}}{3}\\
 \Rightarrow n = 26
$
The total number of terms in the given series is 26.
Hence to find the sum of all the terms of the series, substitute the values of a, d and n in the formula of sum then we get,
$
{S_n} = \dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right)\\
 = \dfrac{{26}}{2}\left( {2 \times 25 + \left( {26 - 1} \right)3} \right)
$
Solving the equation by using the BODMAS method hence first we have to open all the brackets then,
$
{S_n} = \dfrac{{26}}{2}\left( {2 \times 25 + \left( {25} \right)3} \right)\\
 = \dfrac{{26}}{2}\left( {2 \times 25 + 25 \times 3} \right)
$
In the bracket both term have 25 in multiple so taking 25 as common then add 2 and 3,
$
{S_n} = \dfrac{{26}}{2}\left( {25\left( {2 + 3} \right)} \right)\\
 = \dfrac{{26}}{2}\left( {25 \times 5} \right)\\
 = 13 \times 125\\
 = 1625
$

Hence the sum of the given series $25 + 28 + 31 + ..... + 100$ is 1625.

Note: This summation formula is applicable for the series which have equal difference between their consecutive terms. The series which has the same difference between its all consecutive terms is called arithmetic series.