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

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The sum of the first 110 terms in the given arithmetic progression is -110. This conclusion is derived from the provided sums of the first 10 terms (S(10) = 100) and the first 100 terms (S(100) = 10). The calculations reveal that the common difference (d) is -0.22, leading to the final sum formula S(110) = 55(a1 + a100 - 2.2), which simplifies to -110. The arithmetic progression's first term (a1) and the 110th term (a110) were calculated based on the established relationships.

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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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