Question

Find a formula for the nth term, Tn, of an arithmetic sequence where the 5th term is 31, and the 9th term is 55. Hence, or otherwise, find the 40th term of the sequence.

Answer 1

T₄₀ = 241

Explanation:

the nth term of an arithmetic sequence is

`T_(n)` = a₁ + d(n - 1)

a₁ is the first term and d the common difference

given a₅ = 31 and a₉ = 55, then using the nth term formula

a₁ + 4d = 31 → (1)

a₁ + 8d = 55 → (2)

subtract (1) from (2) term by term on both sides to eliminate a₁

(a₁ - a₁ ) + (8d - 4d) = 55 - 31

0 + 4d = 24

4d = 24 ( divide both sides by 4 )

d = 6

substitute d = 6 into either of the 2 equations and solve for a₁

substituting into (1)

a₁ + 4(6) = 31

a₁ + 24 = 31 ( subtract 24 from both sides )

a₁ = 7

Then nth term formula is

`T_(n)` = 7 + 6(n - 1) = 7 + 6n - 6 = 6n + 1

now use this formula to find T₄₀

T₄₀ = 6(40) + 1 = 240 + 1 = 241