Question

The first Three terms of an arithmetic progression are x,(3x+1), and (7x-4).find the 1)the value of x. 2)10th term.

Answer 1

In an arithmetic progression (AP), the difference between consecutive terms is constant. Let’s denote the first three terms as follows:

- First term: \( a_1 = x \)
- Second term: \( a_2 = 3x + 1 \)
- Third term: \( a_3 = 7x - 4 \)

Since it's an AP, the difference between the second term and the first term should be equal to the difference between the third term and the second term:

\[
a_2 - a_1 = a_3 - a_2
\]

Substituting the expressions for the terms:

\[
(3x + 1) - x = (7x - 4) - (3x + 1)
\]

Simplifying both sides:

1. Left side:
\[
3x + 1 - x = 2x + 1
\]

2. Right side:
\[
7x - 4 - 3x - 1 = 4x - 5
\]

Now, we set the two sides equal to each other:

\[
2x + 1 = 4x - 5
\]

Rearranging gives:

\[
1 + 5 = 4x - 2x
\]
\[
6 = 2x
\]

Now, dividing both sides by 2 gives:

\[
x = 3
\]

Now that we have the value of \( x \), we can find the 10th term of the AP.

### Finding the 10th Term:

The first term \( a_1 \) is \( x = 3 \).

To find the common difference \( d \):
\[
d = a_2 - a_1 = (3x + 1) - x = 2x + 1
\]
Substituting \( x = 3 \) into \( d \):
\[
d = 2(3) + 1 = 6 + 1 = 7
\]

Now, the formula for the nth term of an AP is given by:
\[
a_n = a_1 + (n - 1)d
\]
For the 10th term (\( n = 10 \)):
\[
a_{10} = a_1 + (10 - 1)d = 3 + 9 \times 7
\]
Calculating further:
\[
a_{10} = 3 + 63 = 66
\]

### Answers:
1. The value of \( x \) is \( 3 \).
2. The 10th term is \( 66 \).