Question

The first three terms of an A.P are x, (3x+1), (7x-4) .find (1) the value of a
(2) 10th term

Answer 1

In A.P

a = initial term

d = common difference

n-th term:

an = a + ( n - 1 ) d

a1 = a

a2 = a + d

a3 = a + 2 d

In this case

a1 = x

a2 = 3 x + 1

a3 = 7 x - 4

so:

a1 = x

a2 = a1 + d

3 x + 1 = x + d

a3 = a1 + 2 d

7 x - 4 = x + 2 d

Now you must solve system:

3 x + 1 = x + d

7 x - 4 = x + 2 d

First equation.

3 x + 1 = x + d

Subtract x to both sides.

2 x + 1 = d

d = 2 x + 1

Second equation.

7 x - 4 = x + 2 d

Subtract x to both sides.

6 x - 4 = 2 d

2 d = 6 x - 4

Divide both sides by 2.

d = 3 x - 2

Now:

d = d

2 x + 1 = 3 x - 2

Subtract 2x to both sides.

1 = x - 2

Add 2 to both sides.

3 = x

x = 3

d = 2 x + 1

d = 2 • 3 + 1 = 6 + 1

d = 7

Or

d = 3 x - 2

d = 3 • 3 - 2 = 9 - 2

d = 7

1)

a = a1 = x

a = 3

2)

a10 = a1 + 9 d

a10 = 3 + 9 ∙ 7 = 3 + 63 = 66

Your A.P.

3 , 10 , 17 , 24 , 31 , 38 , 45 , 52 , 59 , 66 ...