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raketka [301]
3 years ago
5

The nth term of this sequence is ansquared+by+c 1,11,27,49 Find the values of a b and c

Mathematics
1 answer:
stiv31 [10]3 years ago
4 0

Answer:

The values of a, b and c are:

  • a = 3
  • b = 1
  • c = -3

Step-by-step explanation:

Step 1:

First Confirm the sequence is quadratic by finding the second difference.

Sequence = 1,11,27,49

1st difference:

  • 11-1=10
  • 27-11=16
  • 49-27=22

2nd difference:

  • 16-10=6
  • 22-16=6

Step 2:

Just divide the second difference by 2, you will get the value of a.

6 ÷ 2 = 3

So the first term of the nth term is 3n^{2}

Step 3: Next, substitute the number 1 to 5 into  3n^{2}

n = 1,2,3,4,5

3n² = 3,12,27,48,75

Step 4:

Now, take these values (3n²) 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

3n² = 3,12,27,48,75

Differences:

1 - 3 = -2

11 - 12 = -1

27 - 27 = 0

49 - 48 = 1

Now the nth term of these differences (-2,-1,0,1) is (n - 1) - 2.

so b = 1, and c = -3

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

3n² + (n - 1) -2

= 3n² + n - 1 - 2

= 3n² + n - 3

Therefore, the values of a, b and c are:

  • a = 3
  • b = 1
  • c = -3
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