Answer:
[tex]\sf S_n = \dfrac{n}{2}[2a + (n - 1)d] [/tex]
Step-by-step explanation:
The sequence provided is an arithmetic sequence where [tex]\sf a [/tex] is the first term and [tex]\sf d [/tex] is the common difference between consecutive terms.
The [tex]\sf n [/tex]th term of an arithmetic sequence can be expressed as:
[tex]\sf T_n = a + (n - 1)d [/tex]
To find the sum of the first [tex]\sf n [/tex] terms of the sequence, denoted as [tex]\sf S_n [/tex], we can use the formula for the sum of an arithmetic series:
[tex]\sf S_n = \dfrac{n}{2}(a + T_n) [/tex]
However, since [tex]\sf T_n = a + (n - 1)d [/tex], we substitute it into the formula:
[tex]\sf S_n = \dfrac{n}{2}(a + a + (n - 1)d) [/tex]
[tex]\sf S_n = \dfrac{n}{2}(2a + (n - 1)d) [/tex]
[tex]\sf S_n = \dfrac{n}{2}(2a + nd - d) [/tex]
[tex]\sf S_n = \dfrac{n}{2}(2a + nd - d) [/tex]
[tex]\sf S_n = \dfrac{n}{2}(2a + (n - 1)d) [/tex]
[tex]\sf S_n = \dfrac{n}{2}[2a + (a + (n - 1)d) - a] [/tex]
[tex]\sf S_n = \dfrac{n}{2}[2a + a + (n - 1)d - a] [/tex]
[tex]\sf S_n = \dfrac{n}{2}[3a + (n - 1)d] [/tex]
Thus, the sum of the first [tex]\sf n [/tex] terms of the sequence [tex]\sf a, a + d, a + 2d, a + 3d, \ldots [/tex] is given by the formula:
[tex] \Large\boxed{\boxed{\sf S_n = \dfrac{n}{2}[2a + (n - 1)d] }}[/tex]
Answer:
[tex]S_n=\dfrac{n}{2}\left[\:2a+(n-1)d\:\right][/tex]
Step-by-step explanation:
Given sequence:
[tex]a, \;a + d, \;a + 2d, \; a + 3d, ...[/tex]
A series is the sum of the terms in a sequence, so we can express the series of the first n terms as:
[tex]S_n=a+(a+d)+(a+2d)+...+(a+(n-3)d)+(a+(n-2)d)+(a+(n-1)d)[/tex]
Reverse the order of the terms:
[tex]S_n=(a+(n-1)d)+(a+(n-2)d)+(a+(n-3)d)+...+(a+2d)+(a+d)+a[/tex]
Now, add the two expressions.
[tex]2S_n=(2a+(n-1)d)+(2a+(n-1)d)+(2a+(n-1)d)+...+(2a+(n-1)d)[/tex]
So we’ve now got the term (2a + (n - 1)d) repeated n times, which is:
[tex]2S_n=n\cdot (2a+(n-1)d)[/tex]
Divide both sides by 2:
[tex]S_n=\dfrac{n}{2}\left[\:2a+(n-1)d\:\right][/tex]
Therefore, the sum of the first n terms of the sequence is:
[tex]\Large\boxed{\boxed{S_n=\dfrac{n}{2}\left[\:2a+(n-1)d\:\right]}}[/tex]
