- #1

Government$

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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