Question

What is the nth term for quadratic sequence 7,14,23,34,47,62,79

Answer 1

To find the nth term of a quadratic sequence, we can follow these steps:

Step 1: Find the first difference.
To do this, we calculate the difference between the consecutive terms of the sequence.

14 - 7 = 7
23 - 14 = 9
34 - 23 = 11
47 - 34 = 13
62 - 47 = 15
79 - 62 = 17

Now we have the first difference sequence: 7, 9, 11, 13, 15, 17.

Step 2: Find the second difference.
If it's a quadratic sequence, the second difference should be constant. That is, the difference of the first difference sequence should be the same. Let's calculate it.

9 - 7 = 2
11 - 9 = 2
13 - 11 = 2
15 - 13 = 2
17 - 15 = 2

Great, our second difference is constant and equal to 2.

Step 3: Create the general quadratic formula.
Quadratic sequences are defined by the formula:
a_n = an^2 + bn + c

Since the second difference is constant and equal to 2, we know that 2a should be equal to the second difference. Therefore, a is half of the second difference.

a = 2/2 = 1
So our sequence has the form:
a_n = n^2 + bn + c

Step 4: Find b and c.
We use the given terms from the sequence to find the values of b and c. We know that the first term (when n=1) is 7.

a_1 = 1^2 + b(1) + c = 7
1 + b + c = 7
b + c = 6 ... [1]

We also know that the second term (when n=2) is 14.
a_2 = 2^2 + b(2) + c = 14
4 + 2b + c = 14
2b + c = 10 ... [2]

At this point, we have two equations with two variables. We can resolve this system of equations to find b and c.

Step 5: Solve the system of equations.
We can subtract equation [1] from equation [2] to find the value of b.

(2b + c) - (b + c) = 10 - 6
2b - b + c - c = 4
b = 4

Plug the value of b into either equation [1] or [2] to find c.

b + c = 6
4 + c = 6
c = 6 - 4
c = 2

Step 6: Write down the nth term formula with the found coefficients a, b, and c.
a_n = n^2 + 4n + 2

Therefore, the nth term of the quadratic sequence 7, 14, 23, 34, 47, 62, 79 is a_n = n^2 + 4n + 2.

Answer 2

The n-th term for the sequence will be:
`n^2+4n+2`

Explanation:

Given sequence is: 7, 14, 23, 34, 47. 62, 79, ........

The n-th term of a quadratic sequence is:
`t_(n)=an^2 +bn+c`

For
`n=1`....

`t_(1)=a(1)^2+b(1)+c\\ \\ a+b+c=7 .............................(1)`

For
`n=2`....

`t_(2)=a(2)^2+b(2)+c\\ \\ 4a+2b+c=14 .............................(2)`

For
`n=3`....

`t_(3)=a(3)^2+b(3)+c\\ \\ 9a+3b+c=23 .............................(3)`

Subtracting equation (1) from equation (2), we will get......

`3a+b=7..........................(4)`

Subtracting equation (2) from equation (3), we will get.......

`5a+b=9..........................(5)`

Now, subtracting equation (4) from equation (5)...........

`2a=2\\ \\ a=(2)/(2)=1`

Plugging this
`a=1` into equation (4), we will get....

`3(1)+b=7\\ \\ 3+b=7\\ \\ b=7-3=4`

Now, plugging
`a=1` and
`b=4` into equation (1) .........

`1+4+c=7\\ \\ 5+c=7\\ \\ c=7-5=2`

Thus, the n-th term for the sequence will be:
`n^2+4n+2`