Calculate S(68) for the arithmetic sequence a(n) = (2/3)n + 1/6
87/2
273/6
139/3
4,726/3
The standard formula for an A.P. is:
a(n) = a(1) + (n-1)d
=> a(n) = (2/3)n + 1/6
= 5/6 - 4/6 + (2/3)n
= 5/6 - (n-1)(2/3)
Hence, a(1) = 5/6
d = 2/3
Sum formula:
S(n) = (n/2)[2a(1) + (n-1)d]
Plug in the values to get the sum.
I get the last one: 4,726/3
Well, S(68) represents the sum of the arithmetic sequence from a(1) to a(68). To calculate that, we can use the formula for the sum of an arithmetic sequence: S(n) = (n/2)(a(1) + a(n)).
To find a(1), we substitute n = 1 into the given formula: a(1) = (2/3)(1) + 1/6 = 2/3 + 1/6 = 4/6 = 2/3.
Now, let's find a(68). Plugging n = 68 into the formula, we have a(68) = (2/3)(68) + 1/6 = 136/3 + 1/6 = 272/6 + 1/6 = 273/6.
Now, we're ready to calculate S(68). Plugging in a(1) and a(68) into the sum formula, we get:
S(68) = (68/2)(a(1) + a(68)) = (68/2)(2/3 + 273/6) = 34(2/3 + 273/6) = 34(8/6 + 273/6) = 34(281/6) = (34 * 281) / 6.
Doing the multiplication, we get: (34 * 281) / 6 = 9534 / 6 = 4,726/3.
So, S(68) is equal to 4,726/3.
Just like juggling numbers, that's quite a trick, huh?
To calculate S(68) for the arithmetic sequence a(n) = (2/3)n + 1/6, we need to find the sum of the first 68 terms in the sequence.
The formula to find the sum of arithmetic sequence is given by: S(n) = (n/2)(2a + (n-1)d) where a is the first term and d is the common difference.
In this case, the first term a = (2/3)(1) + 1/6 = 4/6 = 2/3, and the common difference d = (2/3).
Substituting the values into the formula:
S(68) = (68/2)(2(2/3) + (68-1)(2/3))
Simplifying further:
S(68) = (34)(4/3 + 67(2/3))
= (34)(4/3 + 134/3)
= (34)(138/3)
= 4,692/3
Therefore, S(68) for the arithmetic sequence a(n) = (2/3)n + 1/6 is 4,692/3.
To find the sum of the arithmetic sequence, we can use the formula for the sum of an arithmetic sequence, which is given by:
S(n) = (n/2) * (a(1) + a(n))
Where:
- S(n) is the sum of the first n terms of the sequence,
- n is the number of terms in the sequence,
- a(1) is the first term of the sequence, and
- a(n) is the nth term of the sequence.
In this case, we want to calculate S(68) for the arithmetic sequence a(n) = (2/3)n + 1/6.
First, let's find the first term (a(1)) and the 68th term (a(68)):
a(1) = (2/3)(1) + 1/6 = 2/3 + 1/6 = 4/6 + 1/6 = 5/6
a(68) = (2/3)(68) + 1/6 = (136/3) + 1/6 = (136/3) + (2/3) = 138/3
Now we can plug these values into the sum formula to calculate S(68):
S(68) = (68/2) * (5/6 + 138/3)
= (34) * (5/6 + 138/3)
= (34) * (5/6 + 46/1)
= (34) * (5/6 + 46/1)
= (34) * (5/6 + 276/6)
= (34) * (281/6)
= (34 * 281) / 6
= (9534) / 6
= 9,534 ÷ 6
= 4,726/3
Therefore, S(68) for the arithmetic sequence a(n) = (2/3)n + 1/6 is 4,726/3.