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

10/2 (2a+9d) = 60

15/2 (2a+14d) = 165

2a+9d = 12
2a+14d = 22

5d = 10
d=2
so, a = -3

-3 + 17*2 = 31

The kth number kan be written a1+(k-1)d, d difference. The sum of the ten first is a1+a1+d+a1+2d+a1+3d+…+a1+9 and by looking at the a1 and d separately we get 10a1+(1+2+3+…+9)d=10a1+(9*10)/2=10a1+45d=60 using an arithmetic sum. Similarly, we get for the other sum 15a1+(14*15)/2*d=15a1+105d=165. Now divide by 5 and we have the equations 2a1+9d=12 (*) and 3a1+21d=33. Multiplying the first equation by 3 and the other by 2 we get 6a1+27d=36 and 6a1+42d=66. Subtracting of the a1 we obtain 15d=30 or d=2 which in (*) yields a1=-3. Now we can compute the 18th term as a18=a1+17d=-3+17*2=31. Answer: 31

To find the 18th term of the sequence, we need to first determine the common difference of the linear sequence.

Let's denote the first term as "a" and the common difference as "d".

Given that the sum of the first ten terms of the sequence is 60, we can use the formula for the sum of an arithmetic series:

S10 = (n/2)(2a + (n-1)d) = 60

Substituting the given values, we have:

10/2 * (2a + (10-1)d) = 60
5 * (2a + 9d) = 60
10a + 45d = 60

Similarly, the sum of the first fifteen terms of the sequence is 165, so we have:

S15 = (n/2)(2a + (n-1)d) = 165

Substituting the values:

15/2 * (2a + (15-1)d) = 165
7.5 * (2a + 14d) = 165
15a + 105d = 165

Now we have a system of two equations with two variables:

10a + 45d = 60 (equation 1)
15a + 105d = 165 (equation 2)

We can solve this system of equations using substitution or elimination method.

Multiply equation 1 by 3 and equation 2 by -2 to get an equal coefficient for "a":

30a + 135d = 180 (equation 3)
-30a - 210d = -330 (equation 4)

Add equations 3 and 4 to eliminate "a":

-75d = -150
d = 2

Substituting the value of "d" back into equation 1:

10a + 45(2) = 60
10a + 90 = 60
10a = -30
a = -3

Now that we have found the values of "a" and "d" as -3 and 2 respectively, we can use the formula for the nth term of an arithmetic sequence to find the 18th term:

an = a + (n - 1)d

Substituting the values:

a18 = -3 + (18 - 1)(2)
a18 = -3 + (17)(2)
a18 = -3 + 34
a18 = 31

Therefore, the 18th term of the sequence is 31.

To find the 18th term of the sequence, we first need to find the common difference (d) and the first term (a) of the sequence.

Let's start by using the given information. We know that the sum of the first ten terms is 60. This means that:

(a + (a + d) + (a + 2d) + ... + (a + 9d)) = 60 ----> Equation 1

Similarly, the sum of the first fifteen terms is given as 165. This means that:

(a + (a + d) + (a + 2d) + ... + (a + 14d)) = 165 ----> Equation 2

To find the common difference (d) and the first term (a), we can subtract Equation 1 from Equation 2, which gives:

((a + (a + d) + (a + 2d) + ... + (a + 14d)) - (a + (a + d) + (a + 2d) + ... + (a + 9d))) = 165 - 60

Simplifying this equation gives:

(a + 14d - a - 9d) = 105

5d = 105

Now, we can solve for d by dividing both sides of the equation by 5:

d = 105 / 5

d = 21

We have now found the common difference, which is 21.

To find the first term (a), we can substitute d = 21 back into Equation 1 or Equation 2. Let's use Equation 1:

(a + (a + 21) + (a + 2*21) + ... + (a + 9*21)) = 60

Simplifying this equation gives:

(10a + 9d + 8d + ... + 7d) = 60

(10a + 9d + 8d + ... + 7d) = 60

(10a + 21 + 21 + ... + 21) = 60

(10a + 21*9) = 60

10a + 189 = 60

10a = 60 - 189

10a = -129

a = -129 / 10

a = -12.9

Therefore, the first term (a) of the sequence is -12.9 and the common difference (d) is 21.

Now, we can find the 18th term (T18) using the formula for the nth term of an arithmetic sequence:

Tn = a + (n - 1) * d

Substituting the values we have found:

T18 = -12.9 + (18 - 1) * 21

T18 = -12.9 + 17 * 21

T18 = -12.9 + 357

T18 = 344.1

Therefore, the 18th term of the sequence is 344.1.