t1 and t2 are the first two terms of an arithmetic sequence. t1 = 5 and t2 = 7. What is the first term in this sequence that exceeds 2000?

We have the sequence: 5, 7, 9,....

i.e. common difference is 2

nth term is 5 + 2(n - 1) => 2n + 3

We require when 2n + 3 > 2000

i.e. 2n > 1997

so, n > 998.5

As n must be a whole number, n = 999

Hence, 999th term is the first which exceeds 2000

:)>

For an arithmetic sequence:

t₁ = 5

t₂ = t₁ + d ← where d is the common difference → given that: t₂ = 7

7 = t₁ + d

d = 7 - t₁ → given that: t₁ = 5

d = 2

t₃ = t₂ + d = t₁ + 2d

t₄ = t₃ + d = t₁ + 3d

…and you can generalize writing:

t(n) = t₁ + (n - 1).d → the first term in this sequence that exceeds 2000 → t(n) > 2000

t₁ + (n - 1).d > 2000

(n - 1).d > 2000 - t₁ → recall: t₁ = 5

(n - 1).d > 1995

n - 1 > 1995/d → recall: d = 2

n - 1 > 1995/2

n > (1995/2) + 1

n > (1995 + 2)/2

n > 1997/2

n > 998.5 → as n is an integer

n = 999

You can deduce that t₉₉₉ exceeds 2000 → to go further

t(n) = t₁ + (n - 1).d  → when: n = 999

t₉₉₉ = 5 + (999 - 1).2

t₉₉₉ = 2001