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.

Answer 1

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

Answer 2

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