Vincen wrote: ↑

Wed Mar 31, 2021 8:00 am

If the sum of the first five terms of an Arithmetic sequence is equal to 120 and the sum of the next five terms of the same Arithmetic Sequence is equal to 245, what is the 4th term of this Sequence?

A) 29
B) 34
C) 81
D) 86
E) 91

Answer: A

Solution:

We can let the first term = x and the common difference = d. Thus, the second term = x + d, the third term = x + 2d, and so on. We can create the equations:

x + (x + d) + (x + 2d) + (x + 3d) + (x + 4d) = 120 → 5x + 10d = 120

and

(x + 5d) + (x + 6d) + (x + 7d) + (x + 8d) + (x + 9d) = 245 → 5x + 35d = 245

Subtracting the first equation from the second equation, we have:

25d = 125

d = 5

Substituting d = 5 into the first equation, we have:

5x + 50 = 120

5x = 70

x = 14

Since the fourth term is x + 3d, the fourth term is 14 + 3(5) = 29.

Answer: A