Algebra2 Word Problems? Check them? ASAP? 10 points?

a_n = a_1 + (n-1)d

a_1 = 21000, d = 1300 and n = 24 is not correct as in 23 years is asked n = 23

and it will be

a_23 = 21000 +24(1300) = 50900

Tn = a + (n – 1)d

T23 = 21000 + (23 – 1)(1300)

= 21000 + (22)(1300)

= 49600

series is

4, 10, 16

a = 4, d = (10 – 4) = (16 – 10) = 6

n = 25

Tn = a + (n – 1)d

T25 = 4 + (25 – 1)(1300)

= 4 + (24)(6)

= 148

S25 = a/2[2a + (n – 1)d]

= 25/2 [2 × 4 + (25 – 1)(6)]

= 25/2 [8 + (24)(6)]

= 25/2 (152)

= 1900

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You are close, and I can see why you are confused, because the algebra and the question are not explained clearly.

The way I read the question, n should be 23 years because the initial cost starts at year 0 (i.e. before any time is counted). so a_n = a_0 + 23d. If n was 24, you are saying 23 years after 1 year has already past, which does not appear to be the case.

Alternatively, you could say a_1 = 22300 (21000 + 1300), which gives the equation:

a_n = 22300 + 23 * 1300. This is the same as saying a_1 is "the initial cost, plus the cost of the first year"

When you get a question like this, don't be afraid to underline parts of the question and justify your working to illustrate that you've answered the question as it was asked. Teachers sometimes get it wrong, but won't admit it because they don't want to lose credibility. Rebel against them I say!

For the blocks, the pattern is that 6 extra blocks are added for each level. In this case you DO start at n = 1, since a_1 = 4.

This gives:

a_n = a_1 + d(n-1)

= 4 + 6(25-1)

= 4 + 6*25 - 6

= -2 + 150

= 148

This is the same as saying the 0th level is -2, but it doesn't make sense from a practical perspective since you can't start with -2 blocks.

Edit: I didn't read the question properly, and 148 is the base layer of blocks for the pyramid.

One way of finding the average is to find the value for the 13th layer in this case (the half way mark and thus the average number of blocks per layer). Multiply this by the number of layers (25).

E.g. 4, 10 and 16 blocks would give an average of 10 blocks per layer, and multiplying by 3 gives 30 (= 4 + 10 + 16)

a_13 = 4 + 6 * 13 = 82 (average)

82 * 25 = 2050

2050 blocks in total.

1).

A tuition scale for a state university shows an increase of $1,300 each year.

If the initial cost was $21,000.

what will the cost of tuition be in twenty three years?

Arithmetic Sequence

nth or General Term of an Arithmetic Sequence

an = a1 + (n - 1)d

where a1 is the first term of the sequence and d is the common difference.

a1 = 21000, d = 1300 and n = 24

a24 = 21000 + (24 - 1)(1300) = 50900

2).

A child is creating a pyramid with building blocks.

The top three levels include 4 blocks, 10 blocks, and 16 blocks.

How many blocks would be needed for a pyramid 25 levels tall?

Arithmetic Series

Tn = a + (n – 1)d

T25 = 4 + (25 – 1)(6)

= 148

The Sum of the First n Terms of an Arithmetic Sequence

Sn = n/2(a1 + an)

a1 is the first term of the sequence and an is the nth term of the sequence.

Putting in 4 for the first term, 148 for the last term, and 25 for n, we get:

Sn = n/2(a1 + an)

S20 = 25/2(4 + 148) = 1900

Total = 1900 blocks

In question 1 your 2nd solution is correct for sure..

In first year it is 21000=a

Increase=d=1300

T23=a plus 22d

In next question,

Top block has 4=a

2nd top block has 10 and so on... So d=10-6=4

For 25 levels=n,

But here you need to add up all the blocks required,

So by using summation formula of a.p. I.e.

S25=n/2(2a plus (n-1)d)

Putting values,

No of blocks=1900

Hope that helpd!

"in 23 years" implies "at the beginning of the twenty third year" not at the end.

Therefore, only 22 years would have elapsed from now

so 49600 would be the correct answer.common difference = 6

you have 24 common differences after th4 4 blocks

so a25 = 148

152•25/2 =1900

I get 1900