Answer:
a) possible progressions are 5
b) the smallest and largest possible values of the first term are 16 and 82
Explanation:
Sum of terms:
- Sₙ = n/2(a₁ + aₙ) = n/2(2a₁ + (n-1)d)
- S₂₀ = 20/2(2a₁ + 19d) = 10(2a₁ + 19d)
- 2020 = 10(2a₁ + 19d)
- 202 = 2a₁ + 19d
In order a₁ to be an integer, d must be even number, so d = 2k
- 202 = 2a₁ + 38k
- 101 = a₁ + 19k
Possible values of k= 1,2,3,4,5
- k = 1 ⇒ a₁ = 101 - 19 = 82
- k = 2 ⇒ a₁ = 101 - 38 = 63
- k = 3 ⇒ a₁ = 101 - 57 = 44
- k = 4 ⇒ a₁ = 101 - 76 = 25
- k = 5 ⇒ a₁ = 101 - 95 = 16
As per above,
- a) possible progressions are 5
- b) the smallest and largest possible values of the first term are 16 and 82