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# Find the first six terms of the sequence. a_{1} = 7, a_{n} = a_{n - 1} + 6

**Solution:**

Given, a_{1} = 7

a_{n} = a_{n-1} + 6

We have to find the first six terms.

The given series is an arithmetic sequence, with a common difference of 6.

Here, a = 7

\(a_{2} = a_{2-1} + 6\\a_{2} = a_{1} + 6\\a_{2} = 7 + 6\\a_{2} =13\)

\(a_{3} = a_{3-1} + 6\\a_{3} = a_{2} + 6\\a_{3} = 13 + 6\\a_{2} = 19\)

\(a_{4} = a_{4-1} + 6\\a_{4}=a_{3} + 6\\a_{4} = 19 + 6\\a_{2} = 25\)

\(a_{5} = a_{5-1} + 6\\a_{5} = a_{4} + 6\\a_{5} = 25 + 6\\a_{2} = 31\)

\(a_{6} = a_{6-1} + 6\\a_{6} = a_{5} + 6\\a_{6} = 31+6\\a_{2} = 37\)

Therefore, the first six terms of the sequence are 7, 13, 19, 25, 31 and 37.

## Find the first six terms of the sequence. a_{1} = 7, a_{n} = a_{n - 1} + 6

**Summary:**

The first six terms of the sequence a_{1} = 7, a_{n} = a_{n - 1} + 6 are 7, 13, 19, 25, 31 and 37.

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