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Question: How do you find the \(n{\text{th}}\) term of the sequence \(8,12,16,20,24\)?...

How do you find the nthn{\text{th}} term of the sequence 8,12,16,20,248,12,16,20,24?

Explanation

Solution

Here we are given the arithmetic progression or we say AP which refers to the sequence in which the common difference is there between any two consecutive terms as here it is 128=1612=2016=412 - 8 = 16 - 12 = 20 - 16 = 4.
Here nthn{\text{th}}of the AP is given by Tn=a+(n1)d{T_n} = a + \left( {n - 1} \right)d and aa is first term and dd is common difference.

Complete step by step solution:
Here we are given the sequence whose nthn{\text{th}} term is to be found. Here we have the sequence which is given as 8,12,16,20,248,12,16,20,24 and here we know that it is Arithmetic Progression or we say it as AP.
This can be known as here every two consecutive terms of the sequence have the same difference between the numbers like here we have 128=1612=2016=412 - 8 = 16 - 12 = 20 - 16 = 4.
So every two consecutive terms have the same common difference which is equal to 44
Also we have the first term as 88
We must know that when we need to find the nthn{\text{th}} term of any sequence which is AP we need to remember the formula of the nthn{\text{th}} term which is given as:
Tn=a+(n1)d{T_n} = a + \left( {n - 1} \right)d
Here aa is first term and dd is common difference
So we know that first term is 88 and common difference is 44
Hence we can say that:
a=8 d=4  a = 8 \\\ d = 4 \\\
So we can put these values in the above formula we will get:
Tn=a+(n1)d{T_n} = a + \left( {n - 1} \right)d
Tn=8+(n1)4=8+4n4 Tn=4+4n  {T_n} = 8 + \left( {n - 1} \right)4 = 8 + 4n - 4 \\\ {T_n} = 4 + 4n \\\

Hence we can say that nth term=4+4nn{\text{th term}} = 4 + 4n

Note:
Here if we are told to find the sum up to the nn terms of the sequence we can find it by using the formula which says that:
Sn=n2(2a+(n1)d){S_n} = \dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right)
Or we can also write Sn=n2(a+l){S_n} = \dfrac{n}{2}\left( {a + l} \right) where ll is the last term.