We will find the 41st term of the given arithmetic sequence is equal to 88.
Here we have a sequence:
8, 10, 12, 14, ...
Notice that the difference between two consecutive terms is always 2, then this is an arithmetic sequence.
Thus, the n-th term will be equal to the first term plus (n - 1) times the common difference, which is 2, so we can write it as:
\(A_n = 8 + (n - 1)*2\)
To get the 41st term, we just replace n by 41.
\(A_{41} = 8 + (41 - 1)*2= 8 + 80 = 88\)
Then we can conclude that the 41st term in the sequence is equal to 88.
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