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Homework Statement
Sum of first three members of increasing arithmetic progression is 30 and sum of their squares is 692. What is the sum of the first 15 members?
The Attempt at a Solution
So i have system of equations:
a1 + a2 + a3 = 30
(a1)^2 + (a2^2) + (a3^2) = 692
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3a1 + 3d = 30
3(a1)^2 +6(a2)d + d^2= 691
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d = 10 - a1
and if i plug that in 3(a1)^2 +6(a2)d + d^2= 691 i get
2(a1)^2 - 40a1 -192 = 0
now when i find roots of this equation i get 24 and -4 none of them are correct since it says that at the end of the book that a1 = 10 and d=14. Now i do get d=14 with a1=-4 d=-14 with a=-24 but since it is increasing progression i can rule out d=-14, but apparently my solution is not correct. And my solutions satisfy equation 3a1 + 3d = 30 while theirs doesnt.
P.S. This is how they solved this problem:
let a2=a then we get
3a=30 and 3a^2 + 2d^2=692 from that we get a1=10 and d=14 so S15 = 1410
thank you