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
what is the sum of the first 11 terms of the arithmetic series in which
asub one = -12 and d=5
2 answers
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
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