The sum of the first two terms of an arithmetic progession is 18 and the sum of the next 20 terms is -50.?

Question

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

Who can help me with this maths problem?

Answer 1

Let first term = a
Difference = d

2nd term = a + d

Sum of first 2 terms = a + a + d = 2a + d = 18

Since sum of the next 20 terms is -50, the sum of the first 22 terms is -50 + 18 = -32


S22 = 22/2 (a + a + 21d)
=11(2a + 21d)
= 22a + 231d = -32

We now have 2 equations:

2a + d = 18 --- (i)
22a + 231d = -32 --- (ii)

From (i), d = 18 - 2a

Substitute this in (ii)

22a + 231d = -32
22a + 231(18 - 2a) = -32
22a + 4158 - 462a = -32
-440a = -4190

a = 9.5227
d = -1.045

S100 = 100/2 (9.5227 + 9.5227 + 99*(-1.045))
= -4222.72

Answer 2

Let a[1] be the first term. Then the first term would be a1, and the second term would be a1 + d, so

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

It also follows that the sum of the next 20 terms is -50, so it follows that we know the sum of the first 22 terms to be
(-50 + 18) = -32.

Denote the sum of the first n terms to be S(n). Then

S(22) = -32

However, the sum of an arithmetic sequence is also calculated by:

S(n) = (n/2) (2a1 + (n - 1)d)

Therefore

S(22) = (22/2) (2a1 + (22 - 1)d)
S(22) = 11 (2a1 + 21d)
S(22) = 22a1 + 231d
As stated earlier, S(22) = -32, so we can equate these two.

22a1 + 231d = -32

And, as per the equation earlier,

2a1 + d = 18

Two equations, two unknowns.

I'm not going to show you the details of solving two equations and two unknowns, but the answer you should get for these system of equations

2a1 + d = 18
22a1 + 231d = -32

a1 = 419/44, d = -23/22

Now that we have a1 and d, we can easily solve for the desired sum, S(100)

S(n) = (n/2) (2a1 + (n - 1)d)
S(100) = (100/2) (2[419/44] + (100 - 1)[-23/22])
S(100) = (50) ([419/22] + (99)(-23/22))
S(100) = (10475)/11 + (-2277)/22
S(100) = (20950)/22 - 2277/22
S(100) = 18673/22

Edit: Made an error somewhere. Likely the wrong answer.

Answer 3

Hi,

You know that 18 = 2a + d and -50 = 20/2(2(a + 2d) + d(20-1)).
So 18 = 2a + d and -50 = 10( 2a + 4d + 19d). This simplifies to
18 = 2a + d and -5 = 2a + 23d. Solving this system of equations, you get d = -23/22 and a = 419/44.

So to find the sum of the first 100 terms, you can use

S(sub100) = n/2 ( 2a + d(n - 1)) or
S = 100/2(2*419/44 + (-23/22)(100 - 1)) = -46450/11 which is
-4,222 8/11

Answer 4

S2=2a+d=18
S22=11(2a+21d)=-50+18=-32
solve for a and d and substitute in
S100=50(2a+99d)

Answer 5

2a+d=18----> d=18-2a----->1
11[2a+21d]-18=-50-------->2
subs 1 into 2
22a+231(18-2a)-18=-50
-440a+41958-18=-50
4190=440a
a=9.5227273
d=-1.0455

Answer 6

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