The sum of the first 20 terms of an arithmetic progression is 50, and the sum of the next 20 terms is -50.?

Question

Find the sum of the first hundred terms of the progression.

Answer 1

S20=10[2a+19d]=50
--------->a=(5-19d)/2
S40-S20=20[2a+39d]-50=-50

20(5-19d)+7800=0
100-380d+7800=0
d=20.7895
a=-195
S100=50[2a+99d]=83407.895=83400

Answer 2

The sum of an arithmatic progression is given by

s= n/2(2a+(n-1)d)

Where s is the sum, n is the number of terms, a is the starting number and d is the difference between two numbers.

We know that the first 20 term makes 50 and the next 20 make -50. Because they are the same sum but opposite numbers, it makes sense that the first twenty are greater than 0 but the next 20 (i.e. 21-40) are all negative. So between term 20 and 21, the sign changes, i.e. we cross 0.

positive negative
^^^^^^^ 0 ^^^^^^^^^
1 -> 20 | 21 -> 40

The first thing we are going to do is solve for d. We are going to reverse the numbers temporarily so term 20 is now term 1, 19 is now 2, etc. Addition is commutative so it should be ok.

So if we use the equation,
Set s = 50, n = 20 and a = 1/2d.

a =1/2 d because the zero comes right in the middle of term 20 and 21. Since d is the difference between any two terms, 1/2 is the distance from 0 to term 20. Since we reversed the terms, that 1/2 is actually the first term. Now the equation

s= n / 2 * (2a + (n - 1) * d)
50 = 20 / 2 * (2 (d / 2) + (20 - 1) * d)
50 = 10 * (d + 19d)
5 = 20d
d = 0.25

Next we plug d back in the equation and get
s= n / 2 * (2a + (n - 1) * d)
50 = 20 / 2 * (2 a + (20 - 1) * 0.25)
50 = 10 * (2a + 19 * 0.25)
5 = 2a + 4.75
0.25 = 2a
a = 0.125

Now since we reversed the terms, that a is actually the 20th term. To get the true first term we do
a + 19 * d = 0.125 + 19 * 0.25 = 4.875.

So now we use the euqation one last time. We are now using non-reversed numbers so we have to flip the sign of d.
s= n / 2 * (2a + (n - 1) * d)
s = 100 / 2 * (2 (4.875) + (100 - 1) * -0.25)
s = 50 * (9.75 + 99 * -0.25)
s = 50 * -15
s = -750

So the sum of the first 100 is -750.

Answer 3

enable a[a million] be the 1st term. Then the 1st term could be a1, and the 2nd term could be a1 + d, so a1 + (a1 + d) = 18 2a1 + d = 18 It additionally follows that the sum of the subsequent 20 words is -50, so it follows that all of us know the sum of the 1st 22 words to be (-50 + 18) = -32. Denote the sum of the 1st n words to be S(n). Then S(22) = -32 despite the fact that, the sum of an arithmetic series is likewise calculated via: S(n) = (n/2) (2a1 + (n - a million)d) as a result S(22) = (22/2) (2a1 + (22 - a million)d) S(22) = 11 (2a1 + 21d) S(22) = 22a1 + 231d As reported in the previous, S(22) = -32, so we are in a position to equate those 2. 22a1 + 231d = -32 And, as consistent with the equation in the previous, 2a1 + d = 18 2 equations, 2 unknowns. i'm unlikely to coach you the main factors of fixing 2 equations and a couple of unknowns, however the respond you ought to get for those gadget of equations 2a1 + d = 18 22a1 + 231d = -32 a1 = 419/40 4, d = -23/22 Now that we've a1 and d, we are in a position to actual resolve for the needed sum, S(one hundred) S(n) = (n/2) (2a1 + (n - a million)d) S(one hundred) = (one hundred/2) (2[419/40 4] + (one hundred - a million)[-23/22]) S(one hundred) = (50) ([419/22] + (ninety 9)(-23/22)) S(one hundred) = (10475)/11 + (-2277)/22 S(one hundred) = (20950)/22 - 2277/22 S(one hundred) = 18673/22 Edit: Made an blunders someplace. possibly the incorrect answer.