Answer:
[tex]S_{15}= 50[/tex]
Step-by-step explanation:
Given
[tex]a_7 = \frac{14}{3}[/tex]
[tex]d = -\frac{4}{3}[/tex]
[tex]n = 15[/tex]
Required
The sum of n terms
First, we calculate the first term using:
[tex]a_n = a + (n - 1)d[/tex]
Let [tex]n = 7[/tex]
So, we have:
[tex]a_7 = a + (7 - 1)d[/tex]
[tex]a_7 = a + 6d[/tex]
Substitute [tex]a_7 = \frac{14}{3}[/tex] and [tex]d = -\frac{4}{3}[/tex]
[tex]\frac{14}{3} = a + 6*\frac{-4}{3}[/tex]
[tex]\frac{14}{3} = a -8[/tex]
Collect like terms
[tex]a =\frac{14}{3} +8[/tex]
Take LCM and solve
[tex]a =\frac{14+24}{3}[/tex]
[tex]a =\frac{38}{3}[/tex]
The sum of n terms is then calculated as:
[tex]S_n = \frac{n}{2}(2a + (n - 1)d)[/tex]
Where: [tex]n = 15[/tex]
So, we have:
[tex]S_n = \frac{15}{2}(2*\frac{38}{3} + (15 - 1)*\frac{-4}{3})[/tex]
[tex]S_n = \frac{15}{2}(2*\frac{38}{3} + 14 *\frac{-4}{3})[/tex]
[tex]S_n = \frac{15}{2}(2*\frac{38}{3} - 14 *\frac{4}{3})[/tex]
[tex]S_n = \frac{15}{2}(\frac{2*38}{3} - \frac{14 *4}{3})[/tex]
Take LCM
[tex]S_n = \frac{15}{2}(\frac{2*38-14 *4}{3})[/tex]
[tex]S_n = \frac{15}{2}(\frac{20}{3})[/tex]
Open bracket
[tex]S_n = \frac{15*20}{2*3}[/tex]
[tex]S_n = \frac{300}{6}[/tex]
[tex]S_n = 50[/tex]
Hence,
[tex]S_{15}= 50[/tex]
