What is the 24th term of the arithmetic sequence where a₁ = 8 and a₉ = 56?

Solution:

Given a₁ = 8 and a₉ = 56

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

a₁ = 8 = a --- (1)

a₉ = a + (9 - 1)d

a₉ = a + 8d = 56 --- (2)

Substituting (1) in (2) we get,

⇒ 8 + 8d = 56

⇒ 8d = 48

⇒ d = 6

a₂₄ = 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 a₁ = 8 and a₉ = 56?

Summary:

The 24th term of the arithmetic sequence where a₁ = 8 and a₉ = 56 is 146.