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

Answer:

$x = 2$

$S_n = 63$

Step-by-step 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:

$S_n = \frac{n}{2} * [a + a_n]$

So,, we have:

$S_n = \frac{6}{2} * [3 + 18]$

$S_n = 3 * 21$

$S_n = 63$