i have no clue how to do this and was wondering if anyone could help me?
three sequential terms in an arithmetic progression have a sum of 24 and the sum of their squares is 210. if the first term of this particular arithmetic progression is 17, find the sum of the first 6 terms.
three sequential terms in an arithmetic progression have a sum of 24 and the sum of their squares is 210. if the first term of this particular arithmetic progression is 17, find the sum of the first 6 terms.
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Sure.
Let a be the middle term. Then the first term is a-d and the third is a+d. *These are first and third in the "three sequential terms" sense, they are probably not a1 and a3.
Sum = 24 = a-d + a + a+d = 3a
a = 8
Sum of squares = 210 = (a-d)^2 + a^2 + (a+d)^2
210 = 3a^2 + 2ad - 2ad + 2d^2
210 = 3(8^2) + 2d^2
18 = 2d^2
d = ±3.
One term ("a") in the sequence is 8, but the first term is 17. So d<0, so d = -3.
Sum(6) = 6(a1 + a6) / 2
a1 = 17
a6 = 17 - 3(6-1) = 2.
Sum(6) = (6/2)*(17+2) = 57.
Let a be the middle term. Then the first term is a-d and the third is a+d. *These are first and third in the "three sequential terms" sense, they are probably not a1 and a3.
Sum = 24 = a-d + a + a+d = 3a
a = 8
Sum of squares = 210 = (a-d)^2 + a^2 + (a+d)^2
210 = 3a^2 + 2ad - 2ad + 2d^2
210 = 3(8^2) + 2d^2
18 = 2d^2
d = ±3.
One term ("a") in the sequence is 8, but the first term is 17. So d<0, so d = -3.
Sum(6) = 6(a1 + a6) / 2
a1 = 17
a6 = 17 - 3(6-1) = 2.
Sum(6) = (6/2)*(17+2) = 57.
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This isn't possible. The square of a positive or negative number will always be positive and the square of 17 is 289, which is already greater than 210. Something isn't right with the information you've given.