The series given is an arithmetic progression consisting of 5 terms with a common difference of 5 and first term 5 → sum{n} = (n/2)(2×5 + (n - 1)×5) = n(5n + 5)/2 = 5n(n + 1)/2 As no terms have been given beyond the 5th term, and the series is not stated to be an arithmetic progression, the above formula only holds for n = 1, 2, ..., 5.
Sum of the first 15 positive integers is 15*(15+1)/2 = 120 Sum of the first 15 multiples of 8 is 8*120 = 960
x is the first term and d is the difference then x + 3d = 15 and sum of first five terms isx + (x+d) + (x+2d) + (x+3d) + (x+4d)so 5x + 10d = 55 ie x + 2d = 11As x + 3d = 15, d = 4 and x = 3,giving the five terms as 3, 7, 11, 15 and 19
The sum of the first 15 odd numbers is 225.
The sum of the first five whole numbers is 10.
The arithmetic mean is an average arrived at by adding all the terms together and then dividing by the number of terms. Example : Add the digits up and then divide the sum by the number of separate numbers. For the numbers 2, 4, and 9, the sum is 15 and the mean is 15/3 or 5.
6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72, 78, 84, 90.
The series given is an arithmetic progression consisting of 5 terms with a common difference of 5 and first term 5 → sum{n} = (n/2)(2×5 + (n - 1)×5) = n(5n + 5)/2 = 5n(n + 1)/2 As no terms have been given beyond the 5th term, and the series is not stated to be an arithmetic progression, the above formula only holds for n = 1, 2, ..., 5.
It is an Arithmetic Progression with a constant difference of 11 and first term 15.
Sum of the first 15 positive integers is 15*(15+1)/2 = 120 Sum of the first 15 multiples of 8 is 8*120 = 960
the sum of the first 15 prime numbers is 328 .
The sum of a geometric sequence is a(1-rn)/(1-r) In this case, a = 8, r = -2 and n=15 So the sum is 8(1-(-2)15)/(1+2) =8(1+32768)/3 =87,384 So the sum of the first 15 terms of the sequence 8, -16, 32, -64.... is 87,384.
x is the first term and d is the difference then x + 3d = 15 and sum of first five terms isx + (x+d) + (x+2d) + (x+3d) + (x+4d)so 5x + 10d = 55 ie x + 2d = 11As x + 3d = 15, d = 4 and x = 3,giving the five terms as 3, 7, 11, 15 and 19
A(1) = 12A(4) = 3 A(10) = -15.
To get the arithmetic mean, sum them up and divide by the count (in this case 10). So we have 12 + 14 + 15 + 16 + 18 + 20 + 21 + 22 + 24 + 25 = 187. Divide by 10 is 18.7
The sum of the smallest 15 positive integers is 120. The sum of the smallest 15 negative integers is -120.
The sum of the first 5 numbers is 15. 1+2+3+4+5=15