A linear sequence has a common difference of 8. Three consecutive terms in the sequence are added together to give a total of 126 find the first three terms

Answer: 34, 42, 50

Step-by-step explanation:

Let the first term be a

Second term will be a+d

Third term will be a+2d

a+a+d+a+2d = 126

3a+3d = 126

Recall that the common difference (d) is 8.

3a+3d = 126

3a + (3*8) = 126

3a+24 = 126

3a = 126 - 24

3a = 102

a = 102/3

a = 34

First term is 34

Second term will be a+d

= 34+8 = 42

Third term will be a+2d

= 34 + (2*8)

= 34+16

= 50

1st term = a = 34

2nd term = a + d = 34+8 = 42

3rd term = a + 2d = 34+2 (8) = 50

= 34,42,50

Step-by-step explanation:

Given;

Common difference d = 8

Sum of three consecutive terms = 126

Using the arithmetic progression AP nth term formula;

nth term = a + (n-1) d

Where;

a = first term

d = common difference

So,

1st term = a

2nd term = a + d

3rd term = a + 2d

Sum of the three consecutive terms;

a + a+d + a+2d = 3a + 3d

Equating to the given value;

3a + 3d = 126

3a + 3 (8) = 126 (d=8 given)

3a = 126 - 24

3a = 102

a = 102/3 = 34

Therefore,

1st term = a = 34

2nd term = a + d = 34+8 = 42

3rd term = a + 2d = 34+2 (8) = 50