Question

If we have a=1, d=2 , n=75 for an A.P. Find $S_n$

Answer 1

Hint: The series in arithmetic progression are in the form of a ,a+d , a+2d,…………a+(n-1)d for n terms.
Sn is the sum of n terms in an arithmetic progression is given by the following formula.
\[{S_n} = \dfrac{n}{2}\left( {2a + (n - 1)d} \right)\]
Where
n= number of terms.
a= First term of an arithmetic progression.
d=Common difference between adjacent terms of an arithmetic progression.
Sn = sum of n terms in an arithmetic progression.
By substituting the values of respective terms and solving it numerically, we get the value of $S_n.$

Complete step-by-step solution:
It is given in the question that
a= First term of an arithmetic progression=1
n= number of terms=75
d=Common difference between adjacent terms of an arithmetic progression=2
By substituting these values in the above mentioned formula, The solution of problem is as follows
\[ {S_n} = \dfrac{n}{2}\left( {2a + (n - 1)d} \right) \\
\Rightarrow {S_n} = \dfrac{{75}}{2}(2 \times 1 + (75 - 1)2) \\
\Rightarrow {S_n} = \dfrac{{75}}{2}(2 + (74)2) \\
\Rightarrow {S_n} = \dfrac{{75}}{2}(2 + 74 \times 2) \\
\Rightarrow {S_n} = \dfrac{{75}}{2} \times 2(1 + 74) \\
\Rightarrow {S_n} = 75 \times (75) \\
\Rightarrow {S_n} = 5625 \]
Hence the sum of n terms in an arithmetic progression starting with term a=1, having a common difference between adjacent terms as d=2 and total number of terms n=75 is 5,625.

Note: Arithmetic progression is a progression whose adjacent terms differ by a common difference d. Arithmetic progression is increasing series if common difference is positive and arithmetic progression is decreasing series if common difference is negative. The nth term of an arithmetic progression is given by tn.
The formula of $t_n$ is given by \[{t_n} = a + (n - 1)d\].
For the arithmetic progression given in the question containing a=1,d=2 ,n=75
\[ {t_n} = 1 + (75 - 1) \times 2 \\
\Rightarrow {t_n} = 1 + 74 \times 2 \\
\Rightarrow {t_n} = 1 + 148 \\
\Rightarrow {t_n} = 149 \]
The nth term of the given A.P is 149