This has to do with arithmetics (sequences and seires). I will give the Best answer to the person who can explain the answer clearly.

Q1: Find k given the consecutive arithmetic sequence:

5,k,k^2-8 (k^2 means k squared)

Q2: Here is a finite arithmetic sequence: 36, 35 and 1/3, 34 and 2/3...,-30 How many terms does the sequence have?

Q3: An arithmetic sequence starts 23, 36, 49, 62. What is the first term of the sequence to exceed 100,000?

## Answers (1)

Keep in mind that the arithmetic sequence is r,r+d,r+2d,r+3d,...

Q1: Find k given the consecutive arithmetic sequence:

5,k,k^2-8 (k^2 means k squared)

Since these are the first three terms.

r=5

r+d=k

5+d=k by substitution

r+2d=k^2-8

5+2d=(5+d)^2-8 by substitution

5+2d=25+10d+d^2-8 By expanding the square.

d^2+8d+12=0 By moving everything to one side.

(d+2)(d+6)=0 So, d=-2 or -6

Try -2 and -6 to see which answer makes sense. Keep in mind that the arithmetic series must strictly increase or strictly decrease.

If d=-2, then k=3, so the first three terms are 5,3,1, so r=5 and d=-2. This makes sense!

If d=-6, then k=-1, so the first three terms are 5,-1,28. This does not strictly increase or strictly decrease so it's not an arithmetic series.

So d=-2 and k=3.

Q2: Here is a finite arithmetic sequence: 36, 35 and 1/3, 34 and 2/3...,-30 How many terms does the sequence have?

r=36

r+d=35 and 1/3

36+d=35 and 1/3 by substitution

d=-2/3.

36+n(-2/3)=-30 where n is the number of additional terms in the sequence.

n=-66(-3/2)=99. So there are 99 terms after the first term. Altogether, there are 100 terms.

Q3: An arithmetic sequence starts 23, 36, 49, 62. What is the first term of the sequence to exceed 100,000?

r=23

r+d=36

23+d=36 By substitution.

d=13

23+13n=100000 where n is the number of terms in the series.

n=999,977/13=76921.3

We must round UP to the nearest integer to make sure we EXCEED 100,000.

23+(13)76922=100,009.