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# What is the 24th term of the arithmetic sequence where a_{1} = 8 and a_{9} = 56?

**Solution:**

Given a_{1} = 8 and a_{9} = 56

Let ‘a’ be the first term and ‘d’ be the common difference of the A.P. then

a_{1} = 8 = a --- (1)

a_{9} = a + (9 - 1)d

a_{9 }= a + 8d = 56 --- (2)

Substituting (1) in (2) we get,

⇒ 8 + 8d = 56

⇒ 8d = 48

⇒ d = 6

a_{24} = a + 23d

= 8 + 23(6)

= 8 + 138

= 146

Therefore, the 24th term of the arithmetic sequence is 146.

## What is the 24th term of the arithmetic sequence where a_{1} = 8 and a_{9} = 56?

**Summary:**

The 24th term of the arithmetic sequence where a_{1} = 8 and a_{9} = 56 is 146.

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