Question

What is the arithmetic nth term for this \[4,10,16,22...\]?

Answer 1

Hint: Before solving this question, the arithmetic progression is defined as a sequence of numbers in order in which the difference of any two consecutive numbers is equal and also a constant value and it is abbreviated as AP. In this problem, we need to find the arithmetic nth term, hence we need to apply the formula for nth term of AP i.e.,
\[{{a}_{n}}=a+\left( n-1 \right)d\], in this which we need to substitute all values in the formula from the given sequence, to get the value of \[{{a}_{n}}\].


Complete step-by-step solution:
The formula is \[{{a}_{n}}=a+\left( n-1 \right)d\]
Here, \[{{a}_{n}}\]is the nth term of the sequence
\[a\]is the first term in the given sequence
 \[n\]is the number of terms
\[d\]is the common difference between any two terms
\[d={{a}_{2}}-{{a}_{1}}\]
\[{{a}_{2}}\]is the second term in sequence
\[{{a}_{1}}\]is the first term in the sequence
Given sequence is \[4,10,16,22...\]
In this, we need to find out the nth term
From the given terms, the first term and second term of the sequence are 10 and 4,
And the common difference is
\[d={{a}_{2}}-{{a}_{1}}\]
\[d=10-4\]
\[\therefore d=6\]
The first term of the sequence is \[a=4\]
Hence to calculate the nth term i.e., \[{{a}_{n}}\], we have the formula as
 \[{{a}_{n}}=a+\left( n-1 \right)d\]
If we substitute the given data we have,
\[\Rightarrow {{a}_{n}}=4+\left( n-1 \right)\times 6\]
Evaluating the terms, we get
\[\Rightarrow {{a}_{n}}=4+6n-6\]
\[\therefore {{a}_{n}}=6n-2\]
Therefore, the value of the nth term is \[{{a}_{n}}=6n-2\].

Note: The common mistake done by the students in arithmetic progression is taking the incorrect formula and taking the incorrect values for last and first term. Students get confused about finding the nth term and n. The common difference will never be calculated according to the difference of greater number from the lesser number.