The largest and smallest of three consecutive terms in an arithmetic sequence differ by 14. half of the smallest term is added to each term and the sum of the resulting three numbers is 120. what is the value of the original smallest term?

22

Step-by-step explanation:

This is late, but we have the first term as x and the last term as x+14. Being that there is 3 terms in the arithmetic sequence, this makes the sequence

x, x + 7, x + 14.

Half of the smallest term is added to each term:

(3/2) x, (3/2) x + 7, (3/2) x + 14.

Summing the terms gives (9/2) x + 21. Solving the resulting equation (9/2) x + 21 = 120 gives

(9/2) x = 99

x = 22.

The value of the original smallest term is 22.

Let the smallest number be x and the common difference be d.

So the three consecutive numbers are

x, x+d, x+2d

And it is given that the difference between smallest and largest is 14.

So we get x+2d-x = 14

2d=14

d=7

So the three consecutive numbers are

x, x+7, x+14.

Half of the smallest number is 0.5x.

Adding it to each number will give x+0.5x, x+7+0.5x, x+14+0.5x

= 1.5x, 1.5x + 7, 1.5x + 14

And there sum is 120.

1.5x + 1.5x + 7 + 1.5x+21 = 120

4.5x + 28=120

4.5x = 92

x = 20 approx.