the first 3 terms of an AP are x, 2x+1 and 5x-1. find x and the sum of the first 10 terms
tn = t1 x (r^n-1)
4x - 3 = (x + 2) x (r^2) / x + 2
4 - 1.5 = r^2
2.5 = r^2
r = 1.58
but I tried testing that by substituting it with equation to try to figure out t1, 2, and 3
4x - 3 = t1 x (1.58^2)
4x - 3 = t1 x (2.5) / 2.5
1.6x - 1.2 = t1
To find x, we can use the fact that the common difference between consecutive terms in an arithmetic progression (AP) is constant.
Let's consider the first two terms: x and 2x+1. The difference between these two terms should be equal to the common difference.
So, (2x+1) - x = x+1.
Now, let's consider the second and third terms: 2x+1 and 5x-1. The difference between these two terms should also be equal to the common difference.
So, (5x-1) - (2x+1) = 2x.
We now have two equations:
x+1 = 2x
2x = (5x-1) - (2x+1)
Solving the first equation, we find:
x = 1
Substituting this value of x into the second equation:
2(1) = (5(1)-1) - (2(1)+1)
2 = (5-1) - (2+1)
2 = 4 - 3
2 = 1
Both equations hold true, so we have found the value of x, which is 1.
Next, to find the sum of the first 10 terms of the AP, we can use the formula:
Sum = (n/2) * (first term + last term)
Here, n represents the number of terms. In this case, n = 10.
To find the last term, we need to find the term that comes after the 9th term, which is the 10th term.
The common difference between consecutive terms is: (2x+1) - x = x + 1.
So, the 10th term = (9th term) + (common difference) = (5x-1) + (x+1) = 6x.
Plugging these values into the sum formula:
Sum = (10/2) * (x + 6x)
Sum = 5 * 7x
Sum = 35x
Since we found that x = 1, we can substitute it into the equation:
Sum = 35(1)
Sum = 35
Therefore, the sum of the first 10 terms of the AP is 35.
To find the value of x, we need to set up an equation using the given terms of the arithmetic progression (AP).
The general form of an AP is: a, a + d, a + 2d, ...
In this case, we have the first three terms:
a = x
a + d = 2x + 1
a + 2d = 5x - 1
To find the common difference (d), we subtract consecutive terms:
(2x + 1) - x = x + 1
(5x - 1) - (2x + 1) = 3x - 2
Since the common difference will be the same for both pairs, we set up the following equation:
x + 1 = 3x - 2
To solve for x, we bring all the x terms to one side of the equation:
x - 3x = -2 - 1
-2x = -3
Dividing both sides by -2:
x = (-3)/(-2) = 3/2 = 1.5
Therefore, x = 1.5.
To find the sum of the first 10 terms of the AP, we can use the formula:
Sn = (n/2)(2a + (n-1)d),
where Sn is the sum of the first n terms, a is the first term, and d is the common difference.
Plugging in the given values:
n = 10 (since we want the sum of the first 10 terms)
a = x = 1.5 (from before)
d = 3x - 2 = 3(1.5) - 2 = 1.5
Substituting these values into the formula:
S10 = (10/2)(2(1.5) + (10 - 1)(1.5))
Simplifying further:
S10 = 5(3 + 9)(1.5)
S10 = 5(12)(1.5)
S10 = 5(18)
S10 = 90
Therefore, the sum of the first 10 terms is 90.