find the sum of the first eight term of a linear sequence whose first term is 6 and whose last term is 46

the sum of the first ten term of a linear sequence is 60 and the sum of the first fifteen term is 165 find the 18 term of the sequence

Just use your formulas that you learned for this topic.

sum of 8 terms = (8/2)(first + last)
= 4(6 + 46) = 208

#2
sum(10) = (10/2)(2a + 9d) = 60
2a + 9d = 12 **
sum(15) = (15/2)(2a + 14d) = 165
2a + 14d = 22 ***

subtract ** from ***
5d = 10
d = 2
sub into **
2a + 18 = 12
a = -3

term(18) = a + 17d = .....

To find the sum of the first eight terms of a linear sequence, you can use the formula for the sum of an arithmetic series:

Sn = (n/2) * (a1 + an)

Where Sn is the sum of the first n terms, a1 is the first term, and an is the nth term.

In this case, the first term (a1) is 6 and the last term (an) is 46. We need to find the sum of the first eight terms, so n = 8.

Using the formula, we can substitute the values:

S8 = (8/2) * (6 + 46)
S8 = 4 * 52
S8 = 208

Therefore, the sum of the first eight terms of the given linear sequence is 208.

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To find the 18th term of a linear sequence, we can use the formula for the nth term (term number n) of an arithmetic sequence:

an = a1 + (n - 1)d

Where an is the nth term, a1 is the first term, n is the term number, and d is the common difference between consecutive terms.

Given that the sum of the first ten terms is 60 and the sum of the first fifteen terms is 165, we can use this information to determine the common difference (d).

Using the sum formula:

Sn = (n/2) * (a1 + an)

For the sum of the first ten terms:

60 = (10/2) * (a1 + a10)
60 = 5 * (a1 + a10)
12 = a1 + a10 ---(1)

Similarly, for the sum of the first fifteen terms:

165 = (15/2) * (a1 + a15)
165 = 7.5 * (a1 + a15)
22 = a1 + a15 ---(2)

Now, we can solve the two equations (1) and (2) simultaneously to find the values of a1 and a10.

To do this, we can subtract equation (2) from equation (1):

12 - 22 = (a1 + a10) - (a1 + a15)
-10 = a10 - a15

Since we know that a15 - a10 is equal to the common difference (d), we can rewrite the equation as:

-10 = d
d = -10

Now that we have the common difference (d), we can substitute it into equation (1) to find the value of a1:

12 = a1 + a10
12 = a1 + (a1 + (-10))
12 = 2a1 - 10
2a1 = 22
a1 = 11

Now that we have a1 and d, we can use the formula for the nth term to find the 18th term:

a18 = a1 + (n - 1)d
a18 = 11 + (18 - 1) * -10
a18 = 11 + 17 * -10
a18 = 11 - 170
a18 = -159

Therefore, the 18th term of the given linear sequence is -159.