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Find the sum sn of the arithmetic sequence a7＝14/3 d＝-4/3 n＝15

Sagot :

MrRoyal

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]

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