What is the Sum of the First 110 Terms in This Arithmetic Progression?

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The discussion revolves around calculating the sum of the first 110 terms of an arithmetic progression, given that the sums of the first 10 and 100 terms are 100 and 10, respectively. The calculations reveal that the first term and the common difference can be derived, leading to a common difference of -0.22. Using these values, the sum of the first 110 terms is determined to be -110. The final conclusion confirms that the calculations are correct. The thread effectively demonstrates the process of solving for the sum in an arithmetic progression.
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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, what is the sum of first 110 terms?

S(10) = (10/2) (a1 + a10) = 100
S(100) = (100/2) (a1 + a100) = 10

a1 + a10 = 20
a1 + a100 = 0.20

a100 = a1 + 99d
a10 = a1 + 9d

i subtracted and got:
a10 - a100 = 19.8
a1 + 9d - a1 - 99d = 19.8
d = -0.22

S(110) = (110/2) (a1 + a110)
a110 = a100 + 10d

S(110) = 55 (a1 + a100 - 2.2) = 55(0.20 - 2.2) = 55(-2) = -110

is this correct?
 
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demonelite123 said:
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, what is the sum of first 110 terms?

S(10) = (10/2) (a1 + a10) = 100
S(100) = (100/2) (a1 + a100) = 10

a1 + a10 = 20
a1 + a100 = 0.20

a100 = a1 + 99d
a10 = a1 + 9d

i subtracted and got:
a10 - a100 = 19.8
a1 + 9d - a1 - 99d = 19.8
d = -0.22

S(110) = (110/2) (a1 + a110)
a110 = a100 + 10d

S(110) = 55 (a1 + a100 - 2.2) = 55(0.20 - 2.2) = 55(-2) = -110

is this correct?
Yes; well done!
 
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