Question

Sum of the first n terms of the arithmicic sequence

Answer 1

First sum : -14,625

Second sum : 11,390

EXPLANATION :

The sum formula of an arithmetic series is :

`S_n=(n)/(2)(a_1+a_n)`

where Sn = sum

n = number of terms

a1 = first term

an = last term

From the problem, we have the series :

`5+0+(-5)+...+(-380)`

We have the first term, a1 = 5 and the last term, an = -380.

But we don't know the number of terms.

Using the nth term formula of an arithmetic series.

`a_n=a_1+d(n-1)`

The difference in the series is 0 - 5 = -5

Let's solve for the value of n :

`\begin{gathered} a_(n)=a_(1)+d(n-1) \\ -380=5-5(n-1) \\ -380-5=-5(n-1) \\ -385=-5(n-1) \\ (-385)/(-5)=n-1 \\ 77=n-1 \\ 77+1=n \\ n=78 \end{gathered}`

So there are 78 terms.

Now use the sum formula :

`\begin{gathered} S_n=(78)/(2)(5-380) \\ S_n=-14625 \end{gathered}`

The sum is -14,625

For the second sum, we have :

`\sum_{n\mathop{=}1}^(85)(3j+5)`

The first term will be :

`3(1)+5=8`

Solve for the last term at j = 85

`3(85)+5=260`

To summarized, we have :

a1 = 8

an = 260

n = 85

Using the sum formula above :

`\begin{gathered} S_n=(85)/(2)(8+260) \\ S_n=11390 \end{gathered}`

The sum is 11,390