Consider the arithmetic sequence 5,7,9,11, …. Derive a formula for (i) the nth term and (ii) the sum to n terms. (iii) Hence find the sum of the first 20 terms.

We can easily identify the first term (5)

The common difference can be found by subtracting the n^th term from the (n+1)^th term

7-5=9-7=11-9=2

Therefore:

U₁=5 and d=2

The IB formula booklet provides the general formula for the nth term and the sum to n terms. Substitute the previously found values into these formulae.

U_n=U₁ + (n-1)d

S_n=ⁿ/₂(2u₁+(n-1)d)

(i)

U_n=U₁ + (n-1)d

Substitute in values of u1 and d

U_n=5+(n-1)2

Simplify the result by expanding brackets

U_n=5+2n-2

U_n=2n+3

(ii)

S_n=ⁿ/₂(2u₁+(n-1)d)

Substitute in values of u₁ and d

S_n=ⁿ/₂(2(5)+(n-1)2)

S_n=ⁿ/₂(10+2n-2)

S_n=5n+n²-n

S_n=4n+n²

(iii)

Substitute n=20 into the formula from (ii)

S_n=4n+n²

S_n=4(20)+(20)²

Solve

S_n=80+400=480