arithmetic sequence problems......?

Question

Determine the first 8 terms of the arithmetic sequence, when the second term is 42 and the 6th term is 72, explain how you determine the terms.

And....

In an arithmetic sequence the 2nd term is 25 and the 11th term is 79, how many terms of the sequence are less than 200? explain

And....

Thanks you very much for the help (^=^)

Answer 1

basically an arithmatic sequence is just an ordered set of numbers where the next number in the sequence is just determined by adding a fixed number to the previous one.

a general formula to determine the nth term (a_n) is just
a_n= a1 + (n - 1)*d

where a1 is the first number in the seqence and "d" is the difference between two consequtive terms.

so now on to your specific problems! the first step is to figure out what a1 and d so you can find any other a_n

we are given a2 = 42 for n=2 _AND_ a6 = 72 for n=6

42 = a1 + (2 - 1)*d = a1 + d
72 = a1 + (6 - 1)*d = a1 + 5*d

to solve for d using the first eqation, solve the second equation for a1 and substitute it back into the first

a1 = 72 - 5*d
42 = (72 - 5*d) + d
42 = 72 - 4*d
d = 30/4 = 15/2 = 7.5
so
a1 = 72 - 5*(7.5) = 34.5
a2 = a1 + d = 42 (check ... same as given in problem)
a3 = a2 + d = 49.5
a4 = a3 + d = 57
a5 = a4 + d = 64.5
a6 = a5 + d = 72 (check ... same as given in problem)
a7 = a6 + d = 79.5
a8 = a7 + d = 87

ok, the second one is similar, except this time you will sovle for a1 and d, but then use the formula to figure out the largest integer n such that a_n < 200.

you are given a2 = 25 (n=2) and a11=79 (n=11) so plugging these into the general formula we get ...
25 = a1 + (2-1)*d = a1 + d
79 = a1 + (11-1)*d = a1 + 10*d
solving for a1 and d as usual ...
a1 = 79 - 10*d
25 = (79 - 10*d) + d
-54 = -9*d
d = 6
a1 = 79 - 10*6 = 19

check we got d and a1 correct !!
a2 = 19 + (2-1)*6 = 25 (good)
a11 = 19 + (11-1)*6 = 79 (good)

now to solve for n ...
a_n = 19 + (n-1)*6

let a_n = 200 and find the whole integer n which is less than or equal to the value we find

200 = 19 + 6*n - 6 = 13 + 6*n
n = 187/6 = 31.167

so the 31st term is the largest term less than 200, so there are 31 terms less than 200.
check that this is correct!!
a_31 = 19 + (31-1)*6 = 199
a_32 = 19 + (32-1)*6 = 218

Answer 2

The formula for the nth term of an arithmetic sequence is
a+(n-1)*d where a is the first term and d the term difference.
Problem 1
a+ (2-1)*d=42, a+(6-1)*d=72. Solving, a=34.50, d=7.50. Hence the sequence is
34.50, 42, 49,50, 57, 64.50, 72. 79.50, 87, 94.50,...
Problem 2
Here, a+d=25, a+10d=79. Hence a=19, d=6.
You have to find term m which satisfies the inequality
19+(m-1)*6<200. This gives m<30.16666
Hence the value for m should be 31, since the 31st term would be 199. Thus there are 31 terms from the first term onwards which are less than 200.

Answer 3

ok
arithmetic sequence means each term is exactly one specific number greater than the previous one. for example, the sequence 2,4,6,8,10 is increasing by 2 for each term. to find the interval of a sequence, you divide the difference between the terms by the number of terms encompassed by the given terms. in the sequence i have given, if we were given that 4 is the 2nd term and 10 is the 5th term, we would divide the difference in the terms (10 - 4 = 6) by the difference in tem number (5 - 2 = 3), so the interval is 2.

also, you can create an equation based upon the zero term and interval once you figure them out.
in the example, the first term is 0(the first term minus the interval), the interval is 2, and with that info, we can create the equation
0(the zero term) + 2(the interval)*x(the number of term) = y (the term of that number)
now, for your problems.
1) 2nd term is 42, and 6th term is 72
72 - 42 = 30
6 - 2 = 4
30/4 = 7.5
so, the first term is (42 - 7.5) = 34.5
and by adding 7.5 to that number we get
1 = 34.5
2 = 42
3 = 49.5
4 = 57
5 = 64.5
6 = 72
7 = 79.5
8 = 87

also, the equation of this problem would be
27(the zero term) + 7.5 * x = y
2) 2nd term is 25, 11th term is 79
79 - 25 = 54
11 - 2 = 9
54 / 9 = 6
so
1 = 19
2 = 25
3 = 31
4 = 37
5 = 43
6 = 49
7 = 55
8 = 61
9 = 67
10 = 73
11 = 79

now that we know the interval is correct, we can use the equation for this sequence (13 + 6x = y), by putting 200 in for y and solving for x. of course we would then round down the number we get. so
13 + 6x = 200
6x = 187
x = 31.3333333333333333333333333333333333
round down to 31, and that is the number of terms in the sequence that are less than 200

if you wish, you can work out each term and see for yourself, but by checking our math using the equation
13 + 6 * 31 = 13 + 186 = 199
we can save some time. the 31st term is 199 and the 32nd would be greater than 200, so there are 31 terms in the sequence that are less than 200.

Answer 4

6*7, 6*8, 6*9, 6*10, 6*11, 6*12

I don't get the 2nd one, I'm too stoned