Question

How many terms of the series -12-10-8.... make the sum equal to 48?
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ANSWER: n=16

Answer 1

Arithmetic Progression/Sequence.
\[S _{n}=a+(n-1)d\]
\[S _{n}=48\]
\[a=-12\] (the first term)
\[d=2\]
(common difference)

Answer 2

Find n.

Answer 3

i know that

Answer 4

i am having problem in simplification

Answer 5

\[Sn=\frac{ n }{ 2 }[2a+(n-1)d]\]

Answer 6

i used this formula

Answer 7

Ah yes sorry about that.

Answer 8

48=n/2 [2(-12) +(n-1)2]

Answer 9

whats next?

Answer 10

\[48=\frac{ n }{ 2 }[2(-12)+(n-1)2]\]

Answer 11

Multiply by 2 to both sides.

Answer 12

When you multiply both side by 2, you get rid of the 2 on RHS(Right Hand Side).

Answer 13

Expand inside the brackets after that.

Answer 14

96=n[2(-12)+(n-1)2]

Answer 15

Yep. Expand the inside first.

Answer 16

so what did you get inside the brackets.
"[..??..]"

Answer 17

96=n[-24+2n-2]

Answer 18

Yep Correct.
Now expand the outside. So just get rid of the brackets.

Answer 19

Collect the like terms.

Answer 20

96=-24n+2n^2-2n

Answer 21

what will be the quadratic

Answer 22

Collect the like terms.

Answer 23

Then move the 96 to the RHS(Right Hand Side).

Answer 24

Show me final equation of the quadratic.

Answer 25

Show me the*

Answer 26

@koli123able

Answer 27

n^2-13n-48=0

Answer 28

Good. Now factorise.

Answer 29

You know the answer is 16

Answer 30

So then it will have this. (x-16)(x+?)=0

Answer 31

n^2-16n-3n-48=0

Answer 32

What are you doing mate?

Answer 33

Factorise and fill in that question mark.
(x-16)(x+?)=0