Given the arithmetic series a1+a2+a3+a4 a 1 + a 2 + a 3 + a 4 8 + 11 + 14 + 17 + . . + 68 what is the value of sum?

Here the first term, a = 8, common difference, d= 11-8= 3 and the last term , l = 68

So before finding the sum, first we need to find the number of terms, and for that we use the formula

a(n) = a+(n-1)d

68 = 8+(n-1)3

Subtracting 8 from both sides

60 = (n-1)3

Dividing both sides by 3

20=n-1

Adding 1 to both sides

n =21

Now we use the formula of sum of n terms of arithmetic series, which is

step-by-step explanation:

always plug in x. in this case, x equals 4.

i'm going to first solve 4(2+3x)

1) 4(2+3(4)) use pemdas to solve.

i'm going to multiply whats in the parenthesis first, so 3 times 4=12.

2)4 (2+12) i'm then going to multiply 4 by both numbers in the parenthesis (because multiplying comes before adding)

4 times 2= 8.   4 times 12= 48

4) (8+48)= 56 (answer)

i'm then going to solve 8+12x and plug in 4 for x.

1) 8 +12(4)

multiplying comes before adding, so i'm gonna multiply 12 and 4.

12 times 4= 48.

2) 8+48=56

both equal the same!

answ9x x 3=

step-by-step explanation: