What are the first three terms of an arithmetic series in which a1=9,an=105,and Sn=741?
a = 9
term(n) = a + (n-1)d = 105
(n-1)d = 105 - a = 105-9 = 96
sum(n) = (n/2)(2a + (n-1)d) = 741
(n/2)(18 + 96) = 741
n/2 = 6.5
n = 13
(n-1)d = 96
12d = 96
d = 8
the terms are 9, 17, 25, ..
Well, if we have an arithmetic series with the first term (a1) as 9 and the last term (an) as 105, it seems we need some "serious" calculations. Let's bring in the ringmaster of math.
So, Sn, which represents the sum of the series, is given as 741. Now, as a "wise" clown, I know that the formula for the sum of an arithmetic series is Sn = n/2 * (a1 + an), where n is the number of terms.
Let's plug in the values we have: 741 = n/2 * (9 + 105). Now, let's do some circus tricks and solve for n.
741 = n/2 * 114. Ah, it seems we need to free n from the clutches of 2, so we'll multiply both sides by 2.
1482 = n * 114. Now, let's divide both sides by 114, and we'll find that n is approximately 13.
Now that we know the number of terms (n), we can find the common difference (d) of the series by subtracting the first term (a1) from the last term (an) and dividing by n - 1. That sounds like an acrobatic calculation, doesn't it?
So, d = (an - a1) / (n - 1) = (105 - 9) / (13 - 1) = 96 / 12 = 8.
With the common difference in our clown pockets, we can now find the first three terms of the series. Drumroll, please!
a1 = 9 (as given)
a2 = a1 + d = 9 + 8 = 17
a3 = a2 + d = 17 + 8 = 25
So, the first three terms of the arithmetic series are 9, 17, and 25. Ta-da!
To find the first three terms of an arithmetic series, we need to use the formulas for the nth term and the sum of the first n terms.
The nth term of an arithmetic series is given by the formula: an = a1 + (n - 1)d
The sum of the first n terms of an arithmetic series is given by the formula: Sn = (n/2)(a1 + an)
Given that a1 = 9, an = 105, and Sn = 741, we can use these formulas to find the values of n and d.
First, let's find the common difference (d):
an = a1 + (n - 1)d
105 = 9 + (n - 1)d
Now, let's find the value of n:
Sn = (n/2)(a1 + an)
741 = (n/2)(9 + 105)
We have two equations:
105 = 9 + (n - 1)d
741 = (n/2)(9 + 105)
Simplifying the first equation:
96 = (n - 1)d
Substituting the value of d from the first equation into the second equation:
741 = (n/2)(9 + 105)
741 = (n/2)(114)
Multiplying both sides by 2:
1482 = n(114)
Dividing both sides by 114:
n = 1482/114
n ≈ 13
Plugging the value of n into the first equation:
105 = 9 + (13 - 1)d
105 = 9 + 12d
96 = 12d
d = 8
Now that we know the common difference (d) is 8, we can calculate the first three terms using the nth term formula:
a1 = 9
a2 = a1 + d = 9 + 8 = 17
a3 = a1 + 2d = 9 + 2(8) = 25
Therefore, the first three terms of the arithmetic series are 9, 17, and 25.
To find the first three terms of an arithmetic series, we need to determine the common difference (d).
First, we can find the number of terms in the series (n) using the formula for the sum of an arithmetic series:
Sn = (n/2) * (a1 + an)
In this case, we are given that Sn = 741, so:
741 = (n/2) * (9 + 105)
Next, we can solve for n:
741 = (n/2) * 114
741 * 2 = 114n
1482 = 114n
n = 1482 / 114
n ≈ 13
So, the number of terms in the series is approximately 13.
To find the common difference (d), we can use the formula:
d = (an - a1) / (n - 1)
Plugging in the given values, we get:
d = (105 - 9) / (13 - 1)
d = 96 / 12
d = 8
Now that we have the common difference, we can find the first three terms of the series:
a1 = 9
a2 = a1 + d = 9 + 8 = 17
a3 = a2 + d = 17 + 8 = 25
Therefore, the first three terms of the arithmetic series are 9, 17, and 25.