Question

Pls solve this for me ryt now wai abeg

...The first three terms of an arithmetic progression (A.P) are (x+1),(4x-2) and(6x-3) respectively .If the last term is 18,find the
a.Value of x b.Sum of the terms of the progression

Answer 1

`x = 2`

`S_n = 63`

Explanation:

Given

`a_1 = x + 1`

`a_2 = 4x -2`

`a_3 = 6x -3`

`a_n = 18`

Solving (a): x

To do this, we make use of common difference (d)

`d = a_2 - a_1`

`d = a_3 - a_2`

So, we have:

`a_3 - a_2 = a_2 - a_1`

Substitute known values

`(6x - 3) - (4x - 2) = (4x - 2) - (x + 1)`

Remove brackets

`6x - 3 - 4x + 2 = 4x - 2 - x - 1`

Collect like terms

`6x - 4x- 3 + 2 = 4x - x- 2 - 1`

`2x- 1 = 3x- 3`

Collect like terms

`2x - 3x = 1 - 3`

`-x = -2`

`x = 2`

Solving (b): Sum of progression

First, we calculate the first term

`a_1 = x + 1`

`a_1 = 2 + 1 = 3`

Next, calculate d

`d = a_2 - a_1`

`d = (4x - 2) - (x +1)`

`d = (4*2 - 2) - (2 +1)`

`d = 6 - 3 = 3`

Next, we calculate n using:

`a_n = a + (n - 1)d`

Where:

`a_n = 18`

`d = 3; a = 3`

So:

`18 = 3 +(n - 1) * 3`

Subtract 3 from both sides

`15 = (n - 1) * 3`

Divide both sides by 3

`5 = n - 1`

Add 1 to both sides

`6 = n`

`n = 6`

The sum of the progression is:

`S_n = (n)/(2) * [a + a_n]`

So,, we have:

`S_n = (6)/(2) * [3 + 18]`

`S_n = 3 * 21`

`S_n = 63`