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Q: What is the sum of the first 15 terms of an arithmetic?

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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.

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 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.

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

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

The sum of the first 15 odd numbers is 225.

Factor the sum of terms as product of the GCC and a sum of 10t+15

The sum of the first five whole numbers is 10.

The sum of the first 15 positive even numbers is 240. (Simply square 15, then add 15 to the result: 15 x 15 = 225. 225 + 15 = 240).

A(1) = 12A(4) = 3 A(10) = -15.

-15

225

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

What is the sum of the first 27 terms of the geometric sequence -3, 3, - 3, 3, . . . ?

The given sequence is an arithmetic progression with common difference d = 4 and first term a = 3.Sum of n terms of an A.P. is given by: Sn = n/2 x [2a + (n-1)d]We need to find sum of 11 terms so n = 11.Putting value of n, a and d we get:S11 = 11/2 x [2x3 + (11 - 1) x 4]S11 = 11/2 x [6 + 40]S11 = 11/2 x 46 = 11 x 23 = 253

There are 5 common differences between seventh and twelfth terms, so the CD is 2.5/5 ie 0.5. First term is therefore 15 - 6 x 0.5 = 12.

The way to find an average (arithmetic mean) of two or more numbers is simple. Simply use the formula below.sum of all the terms in a list / number of terms in the listIn the case of -15 and 20...(-15) + 20 / 2 = 2.5

for the first and second, its 15 and 34

tn=15-3nt1=15-3x1=12t2=15-3x2=9d=t2-t1/d=9-12=-3tn=t1+(n-1)xdtn=12+(12-1)x-312+11x-3=12+(-33)=-21=tnSn=n/2x(tn+t1)=S12=12/2 x(-21 +12)=6 x -9=-54

The sum of 15 and 27 is 42.

15 + w expresses the sum of 15 and w