what is the sum of the first 11 terms of the arithmetic series in which
asub one = -12 and d=5
so a = -12 , d = 5
sum(n) = (n/2)(2a + (n-1)d )
= (11/2)(-24 + 10(5))
= (11/2)(26)
= 143
OR
sum(n) = (n/2)(first + last)
= (11/2)(-12 + 38)
= 143
An = A1 + (n-1)d
average = (A1 + An)/2
sum = average * number of terms
= n (A1 +An)/2
= (n/2) [ 2 A1 + (n-1)d ]
here
= (11/2) [ -24 + (10)5 ]
= (11/2) (26) = 11 * 13 = 143
To find the sum of the first 11 terms of an arithmetic series, you can use the formula:
S = (n/2) * (2a + (n-1)d)
Where:
- S refers to the sum of the series,
- n is the number of terms in the series,
- a is the first term of the series, and
- d is the common difference between the terms.
In this case, n = 11, a(sub one) = -12, and d = 5.
Plugging in these values into the formula, we get:
S = (11/2) * (2(-12) + (11-1)(5))
Let's simplify this expression step by step:
S = (11/2) * (-24 + 10(5))
S = (11/2) * (-24 + 50)
S = (11/2) * 26
To calculate the final answer, multiply 11/2 by 26:
S = 11 * 13
S = 143
So, the sum of the first 11 terms of the arithmetic series is 143.