# Here are the first five terms of a sequence. 30, 29, 27, 24. 20, ... Find, in terms of n, an expression for the nth term of the sequence.How to do this? Please explain clearly...! Thank you

hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted on

30 , 29 , 27, 24, 20 , ...

Let us try and determine the difference between terms.\

a2-a1 = 20 - 30 = -1

a3- a2= 27- 29 = -2

a4- a3= 24- 27 = -3

a5- a4= 20 - 24 = -4

....

We notice that we do not have a constant difference between terms, However, we found that there is a sequence withthe difference between the terms.

Then,

a1= 30

a2= a1 - 1 = a1- 1(0)/2

a3= a2- 2 = (a1-2) -1 = a1- 3 = a1- 3(2)/2

a4= a3-3 = (a1-3) - 3= a1- 6 = a1- 4(3)/2

a5= a4-4 = (a1-6) - 4= a1- 10 = a1- 5(4)/2

a6= a5- 5 = (a1-10) - 5 = a1- 15 = a1- 6(5)/2

......

an = a1 - n(n-1)/2

neela | High School Teacher | (Level 3) Valedictorian

Posted on

The first 5 terms of the sequence are 30, 29,27,24,20.

So a1 = 30, a2 = 29, a3 = 27, a4 = 24, a5 = 20.

a2-a1 = -1, a3-a2 = -2, a4-a3 = -3, a5-a4 = -4...., an-an-1 = n-1.

Therefore an = a1 - 1 -2 -3 - ... -(n-1) = a1- {1+2+3+..(n-1)} = a1-n(n-1)/2.

an = 30-n(n-1)/2.

Therefore, an = 30-n(n-1)/2 is the expression for the nth term in terms of n.

Tally:

a1 = 30 -(1)(0)/2 = 30

a2 = 30 -2(2-1)/2 = 29

a3 = 30 - 3(3-1)/2 = 30 -3 = 27

a4 = 30 - 4(4-1)/2 =  30 - 6 = 24

a5 = 30 -5(5-1)/2 = 30 -10 = 20.