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.

1 answer

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 \( 66 \).