The third & fourth terms of a sequence are 26 & 40. if the second differences are a constant 4, ?

Question

the third and fourth terms of a sequence are 26 and 40. if the second differences are a constant 4, what are the first five terms of the sequence?

Answer 1

Many ways to solve it.

We see the difference is 14 for the third and fourth terms.

If the second difference is 4, then the next difference (5th-4th) is 18.

Therefore the 5th term is 18+40=58.

If 14-x=4, x=10. Therefore, the difference between the 3rd and 2nd is 10.

So, the 2nd term is 26-10=16

Finally, the 1st term and the second term have a difference of 6, because 10-6=4.

So, the first term is 16-6=10

So, the series is:

10,16,26,40,58....

With the differences of:

6,10,14,18...

And second differences of:

2,2,2...

This can also be solved with the general quadratic:

ax^2+bx+c=y

So, if:

r=nth term

s=n+1 th term

t=n-1 th term

then:

(s-r)-(r-t)=4

s-r-r+t=4

s+t-2r=4

That shows that the second difference is a constant 4.

So, using the above method to find the 2nd term 16, we get:

x=2, y=16

x=3, y=26

x=4, y=40

So, after substitution into ax^2+bx+c=y,

4a+2b+c=16

9a+3b+c=26

16a+4b+c=40

And subtraction leads to:

5a+b=10

7a+b=14

And more subtraction to get:

2a=4

a=2

So,

5(2)+b=10

10+b=10

b=0

And,

4a+2b+c=16

4(2)+2(0)+c=16

8+0+c=16

8+c=16

c=8

Therefore:

ax^2+bx+c=y

2x^2+0x+8=y

2x^2+8=y

So,

x=1,

y=2(1)^2+8=2+8=10

x=2,

y=2(2)^2+8=8+8=16

x=3,

y=2(3)^2+8=18+8=26

x=4,

y=2(4)^2+8=32+8=40

x=5,

y=2(5)^2+8=50+8=58

I would use this method because it can be used to find any term, not only the 1st-5th (you can with the first method too, but it will be a much longer process)

One last method i will present:

Divide the second difference by 2 to get a in ax^2+bx+c=y

a=4/2=2

So,

2x^2+bx+c=y

So, at x=3

18+3b+c=26

At x=4,

32+4b+c=40

Subtract.

14+b=14

b=0

and

18+3(0)+x=26

18+c=26

26-18=c

c=8

So,

2x^2+(0x)+8=y

2x^2+8=y

This will also let us get to the equation:

If you have been given the term and not the term number, then:

2(x^2+4)=y

x^2+4=(y/2)

x^2=[y/2]-4

x=+/-sqrt{[y/2]-4}

As we will probably not see a negative term (have you heard of the negative 2nd term?) we will take the positive root.

So,

x=sqrt([y/2]-4)

Also, x must be whole (have you heard of the 1 and a halfth term?), so:

[y/2]-4 can be expressed as an integer x for x^2=[y/2]-4

So, [y/2]-4 is a square number.

So, y must be even for y/2 to be an integer.

Therefore if y/2=n, then:

n-4 is a square number.

So we will reach every square number:

x^2=t for t=n-4

x^2=n-4

So, the values of t resulting in a whole x:

0,1,4,9,16,25....

So, n-4=t, n=4+t, so we get the values of n:

4,5,8,13,20,29...

And, y/2=n, so y=2n, therefore the values of y:

8,10,16, 26, 40, 58....

So, at the (sqrt[t])th term, y=2t+8, and at x=sqrt(t), t=x^2,

so y=2(x^2)+8=2x^2+8=2(x^2+4)

Hope this helped. email for qs. peace