THE FIRST AND LAST TERMS OF AN ARITHMETIC SERIES IS 5 AND 61 RESPECTIVELY WHILE THE SUM OF ALL THE TERMS IS 957. DETERMINE THE NUMBER OF TERMS IN THE SERIES.
If there are n terms,
n/2 (5+61) = 957
n = 29
a = 5
last term = 61
sum(n) = (n/2)(first + last)
957 = (n/2)(5 + 61)
1914 = 66n
n =29
To find the number of terms in the arithmetic series, we can use the formula for the sum of an arithmetic series.
The formula for the sum of an arithmetic series is:
Sum = (n/2) * (first term + last term)
Let's plug in the given information into the formula:
957 = (n/2) * (5 + 61)
Simplifying:
957 = (n/2) * 66
Now we can solve for n:
957 * 2 = 66n
1914 = 66n
Dividing both sides by 66:
n = 1914 / 66
n ≈ 29
Therefore, the number of terms in the series is approximately 29.
To determine the number of terms in the arithmetic series, we can use the following formula:
Sn = (n/2) * (a1 + an)
where:
- Sn is the sum of all the terms
- n is the number of terms
- a1 is the first term
- an is the last term
We are given that the first term, a1, is 5 and the last term, an, is 61. The sum of all the terms, Sn, is 957.
Plugging in these values into the formula, we have:
957 = (n/2) * (5 + 61)
Now, let's simplify this equation:
957 = (n/2) * 66
To get rid of the fraction, we can multiply both sides of the equation by 2:
2 * 957 = n * 66
1914 = 66n
To solve for n, divide both sides of the equation by 66:
1914 / 66 = n
After doing the division, we find that n is approximately 29. In this case, since n represents the number of terms in the series, we can conclude that there are 29 terms in the given arithmetic series.