SOLUTION: If the sum of the first 10 terms and the sum of the first 100 terms of a given arithmetic progression are 100 and 10, respectively, then what is the sum of the first 110 terms?

Question 1191492: If the sum of the first 10 terms and the sum of the first 100 terms of a given arithmetic progression are 100 and 10, respectively, then what is the sum of the first 110 terms?
Found 2 solutions by greenestamps, math_tutor2020:
Answer by greenestamps(13200) (Show Source): You can put this solution on YOUR website!

Let the first term and common difference be a and d.

Sum of first 10 terms: (a)+(a+d)+(a+2d)+...+(a+9d) = 10a+45d
Sum of first 100 terms: (a)+(a+d)+(a+2d)+...+(a+99d) = 100a+4950d

The given sums of the first 10 and first 100 terms give us two equations in a and d:

10a+45d=100
100a+4950d=10

Solve that pair of equations to find first term a and common difference d. Then use those values to find...

Sum of first 110 terms: (a)+(a+d)+(a+2d)+...+(a+109d) = 100a+6050d

You can do the calculations....

Although the values of a and d are "ugly", you should end up with a sum of -110 for the first 110 terms.

Answer by math_tutor2020(3817) (Show Source): You can put this solution on YOUR website!

a = first term
d = common difference

sum of the first 10 terms = a+(a+d)+(a+2d)+...+(a+9d)
sum of the first 10 terms = 10a+45d
Since 1+2+3+..+8+9 = (9/2)*(1+9) = 45
We can form the equation 10a+45d = 100 which I'll refer to as equation M.

sum of the first 100 terms = a+(a+d)+(a+2d)+...+(a+99d)
sum of the first 100 terms = 100a+4950d
Since 1+2+3+..+98+99 = (99/2)*(1+99) = 4950
We can form the equation 100a+4950d = 10 which I'll refer to as equation N.

sum of the first 110 terms = a+(a+d)+(a+2d)+...+(a+109d)
sum of the first 110 terms = 110a+5995d
Note that 1+2+3+...+108+109 = (109/2)*(1+109) = 5995
We can form the equation 110a+5995d = k where k is the answer we're after. I'll call this equation P

We have these equations
M: 10a+45d = 100
N: 100a+4950d = 10
P: 110a+5995d = k

I'll use the elimination method.

If we computed N-10M, then we get
(100a+4950d) - 10(10a+45d) = 10-10(100)
4500d = -990
d = -990/4500
d = -11/50
d = -0.22

Now compute P-11M
(110a+5995d)-11(10a+45d) = k - 11(100)
5500d = k - 1100
k = 5500d+1100
k = 5500(-0.22)+1100
k = -110

Answer: -110

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