Answer :
To find the sum of the first [tex]\( n \)[/tex] terms of an arithmetic progression (A.P.), we use the standard formula for the sum of an A.P. An arithmetic progression is a sequence of numbers in which the difference of any two successive members is a constant. This constant is referred to as the common difference [tex]\( d \)[/tex].
Let's denote the first term of the arithmetic sequence as [tex]\( a \)[/tex].
The sum of the first [tex]\( n \)[/tex] terms of an arithmetic progression is typically denoted as [tex]\( S_n \)[/tex]. The formula for [tex]\( S_n \)[/tex] is derived from the sum of the terms:
[tex]\[ S_n = \frac{n}{2} [2a + (n-1)d] \][/tex]
This formula can be understood step-by-step as follows:
1. First Term and Common Difference:
- The first term (denoted [tex]\( a \)[/tex]) is the initial term of the sequence.
- The common difference (denoted [tex]\( d \)[/tex]) is the difference between successive terms.
2. List the First [tex]\( n \)[/tex] Terms:
- The terms of the A.P. are: [tex]\( a, a+d, a+2d, a+3d, \ldots, a+(n-1)d \)[/tex].
3. Calculate the Sum of the First [tex]\( n \)[/tex] Terms:
- To calculate the sum [tex]\( S_n \)[/tex], we can use the fact that summing the series from both ends towards the middle gives a useful structure. That is:
[tex]\[ S_n = a + (a+d) + (a+2d) + \cdots + [a+(n-1)d] \][/tex]
4. Pairing Terms from the Start and End:
- Writing the sum in reverse order and adding these two sums term by term helps derive the formula:
[tex]\[ S_n = [a+(n-1)d] + (a+ (n-2)d) + \cdots + [a+0d] \][/tex]
Adding these series:
[tex]\[ 2S_n = n[2a+(n-1)d] \][/tex]
5. Solving for [tex]\( S_n \)[/tex]:
- Finally, dividing by 2 to solve for [tex]\( S_n \)[/tex]:
[tex]\[ S_n = \frac{n}{2}[2a + (n-1)d] \][/tex]
6. Confirming the Answer:
- From the options being:
- a. [tex]\(\frac{n}{2}[a+(n-1) d]\)[/tex]
- b. [tex]\(\frac{\pi}{2}[2 a+(n-1) d]\)[/tex]
- c. [tex]\(\frac{n+1}{2}[2 a+(n-1) d]\)[/tex]
- d. [tex]\(\frac{n}{2}[2 a+(a+1) d]\)[/tex]
Based on the correct formula, the right expression is matched in option (a).
Therefore, the correct answer is:
a. [tex]\(\frac{n}{2}[a+(n-1) d]\)[/tex]
Let's denote the first term of the arithmetic sequence as [tex]\( a \)[/tex].
The sum of the first [tex]\( n \)[/tex] terms of an arithmetic progression is typically denoted as [tex]\( S_n \)[/tex]. The formula for [tex]\( S_n \)[/tex] is derived from the sum of the terms:
[tex]\[ S_n = \frac{n}{2} [2a + (n-1)d] \][/tex]
This formula can be understood step-by-step as follows:
1. First Term and Common Difference:
- The first term (denoted [tex]\( a \)[/tex]) is the initial term of the sequence.
- The common difference (denoted [tex]\( d \)[/tex]) is the difference between successive terms.
2. List the First [tex]\( n \)[/tex] Terms:
- The terms of the A.P. are: [tex]\( a, a+d, a+2d, a+3d, \ldots, a+(n-1)d \)[/tex].
3. Calculate the Sum of the First [tex]\( n \)[/tex] Terms:
- To calculate the sum [tex]\( S_n \)[/tex], we can use the fact that summing the series from both ends towards the middle gives a useful structure. That is:
[tex]\[ S_n = a + (a+d) + (a+2d) + \cdots + [a+(n-1)d] \][/tex]
4. Pairing Terms from the Start and End:
- Writing the sum in reverse order and adding these two sums term by term helps derive the formula:
[tex]\[ S_n = [a+(n-1)d] + (a+ (n-2)d) + \cdots + [a+0d] \][/tex]
Adding these series:
[tex]\[ 2S_n = n[2a+(n-1)d] \][/tex]
5. Solving for [tex]\( S_n \)[/tex]:
- Finally, dividing by 2 to solve for [tex]\( S_n \)[/tex]:
[tex]\[ S_n = \frac{n}{2}[2a + (n-1)d] \][/tex]
6. Confirming the Answer:
- From the options being:
- a. [tex]\(\frac{n}{2}[a+(n-1) d]\)[/tex]
- b. [tex]\(\frac{\pi}{2}[2 a+(n-1) d]\)[/tex]
- c. [tex]\(\frac{n+1}{2}[2 a+(n-1) d]\)[/tex]
- d. [tex]\(\frac{n}{2}[2 a+(a+1) d]\)[/tex]
Based on the correct formula, the right expression is matched in option (a).
Therefore, the correct answer is:
a. [tex]\(\frac{n}{2}[a+(n-1) d]\)[/tex]