What is a1 of the arithmetic sequence for which a3=10 and a51=−182?
1 answer:
Answer:
a₁ = 18
Step-by-step explanation:
The n th term of an arithmetic sequence is
= a₁ + (n - 1)d
where a₁ is the first term and d the common difference
Given a₃ = 10 and a₅₁ = - 182, then
a₁ + 2d = 10 → (1)
a₁ + 50d = - 182 → (2)
Subtract (1) from (2) term by term to eliminate a₁
48d = - 192 ( divide both sides by 48 )
d = - 4
Substitute d = - 4 into (1) and evaluate for a₁
a₁ + 2(- 4) = 10
a₁ - 8 = 10 ( add 8 to both sides )
a₁ = 18
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Answer:
(2a, b )
Step-by-step explanation:
Given the endpoints (x₁, y₁ ) and (x₂, y₂ ) then the midpoint is
[ (x₁ + x₂ ),
(y₁ + y₂ ) ]
Here (x₁, y₁ ) = N(2a, 2b) and (x₂, y₂ ) = P(2a, 0), thus
midpoint = [ (2a + 2a),
(2b + 0 ) ] = (2a, b )