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Question
Find:
10th term of the A.P. 1, 4, 7, 10, ...
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Solution
1, 4, 7, 10...
We have:
a = 1
d =\[4 - 1 = 3\]
\[a_{10} = a + (10 - 1)d \left[ a_n = a + \left( n - 1 \right)d \right]\]
\[ = a + 9d \]
\[ = 1 + 9 \times 3 \]
\[ = 28\]
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