Question

Please help me with this arithmetic series question quick? Thanks!
Find the sum: 28 is on top of the E, I=1 is on the bottom of the E, and (8i - 13) is on the right side of the E.
2884
205
3024
211

Answer 1

\(\large \sum \limits_{i=1}^{28} (8i-13)\)

Answer 2

like that ?

Answer 3

Yes :)

Answer 4

remember the "arithmetic sum " formula ?

Answer 5

Sum of n terms in arithmetic sequence, whose first term is \(a\) and last term \(l\) :

\(S_n = \frac{n}{2}(a+l)\)

Answer 6

for the given series, clearly
\(n=28\)

Answer 7

Okay. S I know that n would = 28 but then I'm lost after that

Answer 8

if u can find \(a\) and \(l\),
u can simply use that sum formula

Answer 9

So. A1 would = -5

Answer 10

@Eckwright576 Also remember the generalized formula for sum of an arithmetic series:
S=(n/2)*[2a+(n-1)*d]
where, d is the common difference, a is the first term and n is the no of term.

Answer 11

yes, first term is \(-5\)

Answer 12

how about last term ?

Answer 13

here, last term is 28th term right ?
u need to find \(a_{28}\)

Answer 14

plugin \(i = 28 \) in
\(8i - 13\)

Answer 15

that gives u 28th term

Answer 16

-140?

Answer 17

try again

Answer 18

8(28) - 13

Answer 19

simplify

Answer 20

211

Answer 21

yes !

so u have \(a = -5\)
\(l = 211\)

Answer 22

you're ready to use the sum formula ?

Answer 23

Yeah. I think

Answer 24

good, plug them and see wat u get for sum

Answer 25

I got 2575. But that's none of my answers?

Answer 26

try again, u must have done a calculation mistake

Answer 27

\(S_n = \frac{n}{2}(a+l)\)

\(S_{28} = \frac{28}{2}(-5+211)\)

\(~~~~= 14(206)\)

\(~~~~= ?\)

Answer 28

I figured out my mistake! :) the answer is. 2884

Answer 29

\(\large \color{red}{\checkmark}\)