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# Can anyone help me with my maths work on the nth term ?

Just need someone to go over this for me

### 7 Answers

- Engr. RonaldLv 72 months ago
solving the 1st 4 terms

an = n(n + 1)

a₁ = 1(1 + 1) = 2

a₂ = 2(2 + 1) = 6

a₃ = 3(3 + 1) = 12

a₄ = 4(4 + 1) = 20

The 1st 4 terms are 2,6,12,20

What is the 100th term

a₁₀₀ = 100(100 + 1) = 10,100

Is 53 a term in the sequence: No

Which term is 240

240 = n(n + 1)

240 = n^2 + n

n^2+ n - 240 = 0

(n + 16)(n - 15) = 0

n = - 16, n = 15

The 15th term...

Conclusion

What are the 1st four terms? 2,6,12,20

What is the 100th term? 10,000

Is 53 a term in the sequence? NO.

which term is 240? The 15th term

- PinkgreenLv 72 months ago
T(n)=n(n+1), n>=1=>

T(1)=2; T(2)=6; T(3)=12; T(4)=20; T(100)=10100.

Let n(n+1)=53=>

n^2+n-53=0=>

n=[-1+/-sqr(1+4*53)]/2=>

n=[-1+/-sqr(213)]/2 is not an integer.

Thus, 53 is not in the sequence.

n(n+1)=240=>

n^2+n-240=0=>

n=[-1+/-sqr(1+4*240)]/2=>

n=15 or -16(rejected)

Thus T(15)=240.

- ?Lv 72 months ago
The nth term of a sequence is n(n + 1)

The first four terms are 2, 6, 12, and 20.

The 100th term is 10,100.

Is 53 a term in the sequence? No

240 is the 15th term.

- llafferLv 72 months ago
You are told that:

f(n) = n(n + 1)

What are the first 4 terms? Solve for f(1), f(2), f(3), f(4):

f(1) = 1(1 + 1) and f(2) = 2(2 + 1) and f(3) = 3(3 + 1) and f(4) = 4(4 + 1)

f(1) = 1(2) and f(2) = 2(3) and f(3) = 3(4) and f(4) = 4(5)

f(1) = 2 and f(2) = 6 and f(3) = 12 and f(4) = 20

----

What's the 100th term? Solve for f(100):

f(n) = n(n + 1)

f(100) = 100(100 + 1)

f(100) = 100(101)

f(100) = 10100

----

Is 53 a term? Set f(n) = 53 and solve for n. If n isn't an integer then the answer is no:

f(n) = n(n + 1)

53 = n(n + 1)

53 = n² + n

0 = n² + n - 53

n = [ -b ± √(b² - 4ac)] / (2a)

n = [ -1 ± √(1² - 4(1)(-53))] / (2 * 1)

n = [ -1 ± √(1 + 212)] / 2

n = (-1 ± √213) / 2

213 is not a perfect square so the roots of this quadratic are irrational. Therefore "53" is not a value in the sequence.

----

Which term is 240? Same steps as above except set f(n) = 240:

f(n) = n(n + 1)

240 = n(n + 1)

240 = n² + n

0 = n² + n - 240

n = [ -b ± √(b² - 4ac)] / (2a)

n = [ -1 ± √(1² - 4(1)(-240))] / (2 * 1)

n = [ -1 ± √(1 + 960)] / 2

n = (-1 ± √961) / 2

n = (-1 ± 31) / 2

n = -32/2 and 30/2

n = -16 and 15

Throw out the negative value since n must be an integer greater than zero, so:

n = 15

240 is the 15th term.

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- ?Lv 72 months ago
What are the first 4 terms ?

Take n(n + 1) and substitute n for 1 2 3 and 4

1(1 + 1) = 1 x 2 = 2

2(2 + 1) = 2 x 3 = 6

3(3 + 1) = 3 x 4 = 12

4(4 + 1) = 4 x 5 = 20

The first four terms are 2 6 12 and 20

The 100th term is 100(100 + 1) .. 100 x 101

Which is 10100

Is 53 a term in this sequence ?

NO .. all the terms are even numbers.

Which term is 240 ?

n(n + 1) = 240

15 x 16 = 240

So it is the 15th term.

- ?Lv 72 months ago
To get the first four terms we substitute for n = 1, 2, 3 and 4

i.e. 1(1 + 1) = 2

2(2 + 1) = 6

3(3 + 1) = 12

4(4 + 1) = 20

The 100th term is when n = 100

so, 100(100 + 1) = 10100

All the terms are even, hence 53 being odd is not a term in the sequence.

If the nth term is 240 then we have n(n + 1) = 240

i.e. n² + n = 240

or, n² + n - 240 = 0

so, (n + 16)(n - 15) = 0

Hence, n = 15...i.e. 15th term

:)>

- az_lenderLv 72 months ago
First four terms are 2, 6, 12, 20.

100th term is 10100.

All terms are even numbers, so 53 won't appear.

240 is the 15th term because 15 x 16 = 240.