Question

Here are the first four terms of a quadratic sequence. The nth term of this sequence is an^2 + bn + c. 1 11 27 49.

Find the values of a, b and c.

Can someone plz answer this question I really need this for homework. 15 points available.

Answer 1

Write down the nth term of this quadratic number sequence.

-3, 8, 23, 42, 65...

Step 1: Confirm the sequence is quadratic. This is done by finding the second difference.

Sequence = -3, 8, 23, 42, 65

1st difference = 11,15,19,23

2nd difference = 4,4,4,4

Step 2: If you divide the second difference by 2, you will get the value of a.

4 ÷ 2 = 2

So the first term of the nth term is 2n²

Step 3: Next, substitute the number 1 to 5 into 2n².

n = 1,2,3,4,5

2n² = 2,8,18,32,50

Step 4: Now, take these values (2n²) from the numbers in the original number sequence and work out the nth term of these numbers that form a linear sequence.

n = 1,2,3,4,5

2n² = 2,8,18,32,50

Differences = -5,0,5,10,15

Now the nth term of these differences (-5,0,5,10,15) is 5n -10.

So b = 5 and c = -10.

Step 5: Write down your final answer in the form an² + bn + c.

2n² + 5n -10

U can use it for finding a,b and c

Please mark brain list

Answer 2

Answer:

9+9=18

Step-by-step explanation:

so all u have to do is search up on brainly in