# Find the twentieth and the nth of the sequence 4,9,14,19,....

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4,9,14,19,....a20,...an

We notice that the differenc ebetween terms is 5

==> r= 5

a1= 4

a2= 4+ 5 = 9

a3= 4+ 2(5) = 14

......

a20 = 4 + 19(5) = 99

an = a1 + (n-1)*r

= 4 + (n-1)*5

= 4+ 5n - 5

= 5n -1

==> an = 5n -1

First we have to find the difference between terms:

an = a1 + (n-1)d

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

= (9 - 4) / 1

= 5

To test that this is an arithmetic series we can determine whether the differences are the same for the other two terms:

d = (14 - 4) / 2 = 5

d = (19 - 4) / 3 = 5

So it is an arithmetic series. From here solving for a20 is as simple as using the same expression:

a20 = a1 + (20 - 1)d

a20 = 4 + 19(5)

a20 = **99**

Before any calculus of any term, we'll have to establish if the sequence is an arithmetical progression or geometric progression.

We notice that the difference between 2 consecutive terms is:

9-4 = 5

14-9 = 5

19-14 = 5

.................

So, the sequence is an arithmetical progression where the first term a1 = 4 and the common difference, d = 5.

We'll apply the formula of finding the n-th term of the a.p.

an = a1 + (n-1)*d

**an = 4 + (n-1)*5**

Substituting n by the value 20, we could calculate the 20th term of the a.p.

a20 = a1 + (20-1)*5

a20 = 4 + 19*5

a20 = 4 + 95

**a20 = 99**

same as like this questions,please answer for this .'Find n if nth term of the AP 4,9,14,19...is 24.

The given sequence 4,9,14,19.

T2 -T1 = 9-4 =5.

T3-T2 = 14-9 =5

T4-T3 = 19-14 =5

This is an AP.So the successive terms have the common difference d = 5.

The nth term Tn = T1 + (n-1) d.

n =20

20 th term = T20 = T1 +(20-1)d = 4 +(20-1)*5 = 4+19*5 = 99