Question

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

Answer 1

Answer:

$S_{15}= 50$

Step-by-step explanation:

Given

$a_7 = \frac{14}{3}$

$d = -\frac{4}{3}$

$n = 15$

Required

The sum of n terms

First, we calculate the first term using:

$a_n = a + (n - 1)d$

Let $n = 7$

So, we have:

$a_7 = a + (7 - 1)d$

$a_7 = a + 6d$

Substitute $a_7 = \frac{14}{3}$ and $d = -\frac{4}{3}$

$\frac{14}{3} = a + 6*\frac{-4}{3}$

$\frac{14}{3} = a -8$

Collect like terms

$a =\frac{14}{3} +8$

Take LCM and solve

$a =\frac{14+24}{3}$

$a =\frac{38}{3}$

The sum of n terms is then calculated as:

$S_n = \frac{n}{2}(2a + (n - 1)d)$

Where: $n = 15$

So, we have:

$S_n = \frac{15}{2}(2*\frac{38}{3} + (15 - 1)*\frac{-4}{3})$

$S_n = \frac{15}{2}(2*\frac{38}{3} + 14 *\frac{-4}{3})$

$S_n = \frac{15}{2}(2*\frac{38}{3} - 14 *\frac{4}{3})$

$S_n = \frac{15}{2}(\frac{2*38}{3} - \frac{14 *4}{3})$

Take LCM

$S_n = \frac{15}{2}(\frac{2*38-14 *4}{3})$

$S_n = \frac{15}{2}(\frac{20}{3})$

Open bracket

$S_n = \frac{15*20}{2*3}$

$S_n = \frac{300}{6}$

$S_n = 50$

Hence,

$S_{15}= 50$