Answer:
a₁ = - 11 and d = 3
Explanation:
the sum to n terms of an arithmetic sequence is
=
[ 2a₁ + (n - 1)d ]
where a₁ is the first term and d the common difference
given S₄₀ = 1900 , then
S₄₀ =
[ 2a₁ + 39d ] = 1900 , that is
20(2a₁ + 39d) = 1900 ( divide both sides by 20 )
2a₁ + 39d = 95 → (1)
the nth term of an arithmetic sequence is
= a₁ + (n - 1)d
given a₄₀ = 106 , then
a₁ + 39d = 106 → (2)
subtract (2) from (1) term by term to eliminate d
a₁ + 0 = - 11 , so
a₁ = - 11
substitute a₁ = - 11 into (2) and solve for d
- 11 + 39d = 106 ( add 11 to both sides )
39d = 117 ( divide both sides by 39 )
d = 3