In an arithmetic series t1 = 25, t3= 23.
Find the value of n for which Sn = 135
n/2 (2*25 + (n-1)(23-25)/2) = 135
Well, let's do some calculations to figure this out.
In an arithmetic series, we know that the difference between each term is constant. Let's call this difference "d."
Given that t1 = 25 and t3 = 23, we can find the value of "d" as follows:
d = t3 - t1
= 23 - 25
= -2
Now, let's find the value of "n" for which Sn = 135.
To find the sum of the first "n" terms (Sn) in an arithmetic series, we can use the formula:
Sn = (n/2)(2t1 + (n-1)d)
Substituting the given values into this formula, we can set up the following equation:
135 = (n/2)(2(25) + (n-1)(-2))
To solve this equation, we can simplify it:
135 = (n/2)(50 - 2n + 2)
135 = (n/2)(52 - 2n)
Multiplying both sides by 2 to eliminate the fraction:
270 = n(52 - 2n)
Expanding the right side:
270 = 52n - 2n^2
Rearranging the equation to make it a quadratic equation:
2n^2 - 52n + 270 = 0
Unfortunately, this equation doesn't have simple integer solutions.
To find the value of n for which Sn = 135, we need to find the sum of the arithmetic series and then solve for n.
The formula for the sum of an arithmetic series is:
Sn = n/2 * (t1 + tn)
Given that t1 = 25, we can substitute this value into the formula:
135 = n/2 * (25 + tn)
We also know that t3 = 23, which means the third term of the series is 23.
We can use this information to find tn, the nth term of the series.
The formula for the nth term of an arithmetic series is:
tn = t1 + (n - 1) * d
where d is the common difference between consecutive terms.
Since we know t1 = 25, t3 = 23, and n = 3, we can substitute these values into the formula to solve for d:
23 = 25 + (3 - 1) * d
23 = 25 + 2d
2d = 23 - 25
2d = -2
d = -1
Now that we know d, we can substitute it into the formula for tn and simplify:
tn = 25 + (n - 1) * (-1)
tn = 25 - n + 1
tn = 26 - n
Substituting tn into the formula for Sn and simplifying, we get:
135 = n/2 * (25 + 26 - n)
135 = n/2 * (51 - n)
270 = n * (51 - n)
270 = 51n - n^2
Rearranging the equation, we obtain a quadratic equation:
n^2 - 51n + 270 = 0
Now we can solve this quadratic equation to find the value(s) of n.
Using the quadratic formula, where a = 1, b = -51, and c = 270:
n = (-b ± √(b^2 - 4ac)) / (2a)
Plugging in the values, we get:
n = (-(-51) ± √((-51)^2 - 4 * 1 * 270)) / (2 * 1)
n = (51 ± √(2601 - 1080)) / 2
n = (51 ± √1521) / 2
Taking the square root of 1521, we have:
n = (51 ± 39) / 2
This gives two possible solutions:
n = (51 + 39) / 2 = 90 / 2 = 45
n = (51 - 39) / 2 = 12 / 2 = 6
Therefore, there are two possible values of n for which Sn = 135: n = 45 and n = 6.
To find the value of n in an arithmetic series, we need to know the first term (t1), the common difference (d), and the formula for the sum of an arithmetic series (Sn).
In this case, we are given t1 = 25 and t3 = 23. We can use these values to find the common difference (d).
The formula to find the nth term (tn) in an arithmetic series is: tn = t1 + (n - 1) * d.
Using t1 = 25, we can substitute it into the formula to find t3:
t3 = t1 + (3 - 1) * d
23 = 25 + 2d
23 - 25 = 2d
-2 = 2d
d = -1
Now that we know the common difference (d), we can find the sum of the first n terms (Sn) using the formula: Sn = n/2 * (2t1 + (n - 1) * d).
Substituting the given values, Sn = 135 and t1 = 25, we can solve for n:
135 = n/2 * (2 * 25 + (n - 1) * (-1))
135 = n/2 * (50 - n + 1)
135 = n/2 * (51 - n)
270 = n * (51 - n)
270 = 51n - n^2
n^2 - 51n + 270 = 0
We can solve this quadratic equation by factoring or using the quadratic formula.
The factored form of the quadratic equation is: (n - 6)(n - 45) = 0
Setting each factor to zero, we get two possible values for n: n = 6 or n = 45.
Therefore, the value of n for which Sn = 135 is either 6 or 45.