The sum of the first two terms of an arithmetic progression is 18 and the sum of the first four terms is 52.?

Question

Find the sum of the first eight terms.

Answer 1

a+a+d=18
2a+d=18
a+a+d+a+2d+a+3d=4a+6d=52
2a+3d=26
2d=26-18=8
d=4
2a=14
a=7
sum of the first eight terms = n(2a+(n-1)d)/2 = 8(14+7*4)/2
= 4*(42) = 168

Answer 2

The eight terms are a, b, c, d, e, f, g, and h.
We know that a+b = 18 and c+d = 34 (since a+b+c+d = 52).
To find the terms, I decided to start with trial and error, knowing that there was a pattern to the numbers. My first attempt had terms increasing by 2, but this was too small. My second attempt had terms increasing by 10, but this was too much. I decided to try a difference of 4. This worked.

If a+b = 18 and b = a+4, then a = 7. So we have 7, 11, 15, 19, which satisfies the question. Continuing the progression, the first eight terms are 7, 11, 15, 19, 23, 27, 31, and 35. These terms sum to 168. Your answer is 168.

Answer 3

The sum of the first two terms is 18; therefore, since the first term is a1, and the second term is (a1 + d), then the sum of them would be

a1 + (a1 + d) = 18
2a1 + d = 18

The sum of the first 4 terms is 52, meaning
S(4) = 52, but
S(n) = (n/2) (2a1 + (n - 1)d)
S(4) = (4/2) (2a1 + (4 - 1)d)
S(4) = (2) (2a1 + 3d)
S(4) = 4a1 + 6d = 52

4a1 + 6d = 52 can be simplified by dividing by 2, getting us

2a1 + 3d = 26

Two equations, two unknowns:

2a1 + d = 18
2a1 + 3d = 26

This will yield d = 4, a1 = 7

All we have to do is solve for S(8) using the arithmetic sum formula.

S(n) = (n/2) (2a1 + (n - 1)d)
S(8) = (8/2) (2(7) + (7)(4))
S(8) = (4) (14 + 28)
S(8) = (4) (42) = 168

Answer 4

sum = a million/2N (2a + (n-a million)d) 18 = a million/2 x 2 (2a+(2-a million)d) 18 = 2a+d fifty two = 4x1/2 (2a+(4-a million)d) fifty two = 4a+6d So fifty two=4a+6d 18=2a+d Double the 2nd equation fifty two=4a+6d 36=2a+2nd -------------- sixteen= 4d d = 4 18 = 2a+d 18= 2a+4 14 = 2a A = 7 D = 4 Sum to eight words = 8 x a million/2 (14+(7)x4 = 4 (40 two) = 168

Answer 5

I confirm, 168 with a1= 7 and d = 4

Answer 6

S2=2a+d=18
S4=4a+6d=52
-4a-2d=--36
adding 4d=16
d=4 sub in 2a+4=18
2a=14and a=7
S8=4(2a+7d)
=4(14+28)
=168

Answer 7

Sn=n/2(2a+(n-1)d)
therefore,S2=2/2(2a+(2-1)d)
=2a+d
But S2=18
therefore,2a+d=18..............eqn1

S4=4/2(2a+(4-1)d)
=4a+3d
But S4=52
therefore,4a+3d=52.............eqn 2

multiply eqn1 by 2
= 4a+d=36............................eqn3

sudtract eqn 3 from eqn2
= 2d=16
therefore,d=16/2
=8
substitute d=8 into eqn 1
2a+8=18
2a=10
a=10/2
a=5

therefore,a=5 and d=8

therefore S8=8/2(2(5)+(8-1)8)
=4(10+56)
=4(66)
=264

Answer 8

168 is right answer