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Answer:
a₁₅ = 30
Step-by-step explanation:
The nth term of an arithmetic sequence is
[tex]a_{n}[/tex] = a₁ + (n - 1)d
where a₁ is the first term and d the common difference
Given a₅ = 10 and a₁₀ = 20 , then
a₁ + 4d = 10 → (1)
a₁ + 9d = 20 → (2)
Subtract (1) from (2) term by term to eliminate a₁
5d = 10 ( divide both sides by 5 )
d = 2
Substitute d = 2 into (1) and solve for a₁
a₁ + 4(2) = 10
a₁ + 8 = 10 ( subtract 8 from both sides )
a₁ = 2
Then
a₁₅ = 2 + (14 × 2) = 2 + 28 = 30
Answer:
30
Step-by-step explanation:
Evenly-spaced terms of an arithmetic sequence are an arithmetic sequence. You are given terms 5 and 10 and asked to find term 15. The difference between term 15 and term 10 will be the same as the difference between term 10 and term 5.
5d = a10 -a5 = 20 -10 = 10 . . . . . . . this is 5 times the common difference (d)
a15 = a10 +5d = 20 +10 = 30
The 15th term is 30.
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Additional comment
Another way to look at this is that a10 is the midpoint between a15 and a5. That means ...
a10 = (a15 +a5)/2
a15 = 2(a10) -a5 = 2(20) -10 = 30