I have a bit of problem on the following questions : Sequences and Series

Q

- In a sequence X1,X2,X3.................Xn

X1 =0, X2 = 3, X3= 12 and Xn = a+bn+cn^2

Find the values of a, b, and c

In an arithmetic series, the sum of the first 2n terms is half the sum of the first 3n terms. Given

a= 12, d=3 find n

3.

The sum of three numbers in an arithmetic sequence is 18 and the sum of their squares is 206, find the numbers

Cheers guys!

## Answers (2)

--> a= 3; b=-6; c= 3;

n=7

a+d=6

a^2 + 6^2 + (12-a)^2 = 206

13, 6, -1

1) The problem give 3 equations and 3 unknowns:

a + b + c = 0

a + 2b + 4c = 3

a + 3b + 9c = 12

resolving a=3, b=-6, c=3

2) We know that the sum equals to:

S = an + dn(n-1)/2

then

S2N = S3N / 2

a(2n) + d(2n)(2n-1)/2 = a(3n) + d(3n)(3n-1)/2

12(2n) + 3(2n)(2n-1)/2 = 12(3n) + 3(3n)(3n-1)/2

resolving

n (-3n + 21) = 0

One solution is n=0

The other solution that is better is

-3n + 21 = 0

n = 7

3) a= first number

a+r =second number

a+2r = third number

a + (a+r) + (a+2r) = 18

3a + 3r = 18

a + r = 6 --> note that it is the second number

The other equation is

aÂ² + (a+r)Â² + (a+2r)Â² = 206, resolving and remplacing (a+r=6) then

(a-13)(a+1)=0

a=13 and a=-1, with anycase the result is the same

in a=13 then r=6-a= -7, and the third number is a+2r = -1

Then the three numbers are -1, 6 and 13.