find the sum of the 1st 70 terms of the arithmetic sequence 16,6,-4,-14
The nth term of the sequence is:
T(n)=26-10n, so
T(1)=26-10*1=16,
T(2)=26-10*2=6,
....
T(70)=26-10*70=-674
The sum of an arithmetic sequence is the average of the first and last terms multiplied by the number of terms.
Here, we have
Sum = (T(1)+T(70)*70/2
I'll let you take it from here.
To find the sum of the first 70 terms of the arithmetic sequence with the first term of 16 and a common difference of -10, you can use the formula for the sum of an arithmetic series:
Sn = (n/2) * (2a + (n-1)d),
where Sn represents the sum of the first n terms, a is the first term, d is the common difference, and n is the number of terms.
We know that a = 16, d = -10, and n = 70.
Plugging in these values into the formula, we get:
S70 = (70/2) * (2 * 16 + (70 - 1) * (-10)).
Simplifying further:
S70 = 35 * (32 + 69 * (-10)).
S70 = 35 * (32 - 690).
S70 = 35 * (-658).
S70 = -23030.
Therefore, the sum of the first 70 terms of the arithmetic sequence is -23030.
To find the sum of the first 70 terms of an arithmetic sequence, we need to use the formula for the sum of an arithmetic series. The formula is given by:
Sn = (n / 2) * (2a + (n - 1)d)
Where:
- Sn is the sum of the first n terms
- a is the first term of the arithmetic sequence
- d is the common difference between terms
- n is the number of terms
In this case, the first term (a) is 16, the common difference (d) is -10 (subtracting 10 to each term), and we want to find the sum of the first 70 terms (n = 70).
Substituting these values into the formula, we have:
S70 = (70 / 2) * (2 * 16 + (70 - 1) * -10)
Simplifying further:
S70 = 35 * (32 + 69 * -10)
S70 = 35 * (32 - 690)
S70 = 35 * (-658)
S70 = -23,030
Therefore, the sum of the first 70 terms of the arithmetic sequence is -23,030.