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Find the \[{10^{th}}\] term from the last term of the AP: \[8,10,12.....,126\]

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Hint: Arithmetic Progression (AP) is a sequence of numbers in order in which the difference of any two consecutive numbers is a constant value. For example, the series of natural numbers: \[1,{\text{ }}2,{\text{ }}3,{\text{ }}4,{\text{ }}5,{\text{ }}6,\]is an AP, which has a common difference between two successive terms (say 1 and 2) equal to 1 (2 -1). Even in the case of odd numbers and even numbers, we can see the common difference between two successive terms will be equal to 2.
Definition 1: A mathematical sequence in which the difference between two consecutive terms is always a constant and it is abbreviated as AP.
Definition 2: An arithmetic sequence or progression is defined as a sequence of numbers in which for every pair of consecutive terms, the second number is obtained by adding a fixed number to the first one.
In arithmetic progression (A.P) series the first term is denoted by ‘\[a\]’ and the common difference is denoted by ‘\[d\]’ and ‘\[n\]’ is a number of terms. ‘\[{a_n}\]’ is last term. Here in the series given the value of ‘\[a\]’ is \[8\]and ‘\[d\]’ is\[2\].

Complete step by step answer:
\[8,10,12,....126\]
Here first term \[a\]=\[8\]
Common difference \[ = \]\[10 - 8 = 2\]
Last term (\[l\])\[ = 126\]
Number of terms\[ = 10\]
Now \[{n^{th}}\] term from end using formula
\[l - (n - 1)d\]
\[ = 126 - (10 - 1)2\]
\[ = 126 - 9 \times 2\]
\[ = 108\]

Note: We can also find \[{n^{th}}\] term from beginning by using formula \[{a_n} = a + (n - 1)d\] and also sum of \[n\] terms by using formula \[{S_n} = \dfrac{n}{2}(2a + (n - 1)d)\] or \[{S_n} = \dfrac{n}{2}(a + l)\] if last term is given.