Question

The nth term sequence is n( n - 1) what is the first four terms

What is the 20th Term?

Is 53 apart of the sequence?

Which term is 132?

Answer 1

0,2,6,12

a20=380

no term is 53

n=12

Explanation:

put n=1,2,3,4

a1=1(1-1)=1×0=0

a2=2(2-1)=2×1=2

a3=3(3-1)=3×2=6

a4=4(4-1)=4×3=12

a20=20(20-1)=20×19=380

53=n(n-1)

53=n²-n

n²-n-53=0

`n=(1 \pm√((-1)^2-4(1)(-53)) )/(2* 1) \\=(1 \pm √(1+212))/(2) \\=(1 \pm√(213) )/(2)`

n≈7.797(rejected being a rational number or non integer.)

or

n≈-6.797(Rejected being negative and non integer)

again

132=n(n-1)=n²-n

n²-n-132=0

n²-12n+11n-132=0

n(n-12)+11(n-12)=0

(n-12)(n+11)=0

n=12

or

n=-11(rejected being negative)