Discover answers to your most pressing questions at Westonci.ca, the ultimate Q&A platform that connects you with expert solutions. Explore comprehensive solutions to your questions from knowledgeable professionals across various fields on our platform. Get quick and reliable solutions to your questions from a community of experienced experts on our platform.
Sagot :
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.
_____
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
We hope this was helpful. Please come back whenever you need more information or answers to your queries. Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. Thank you for visiting Westonci.ca, your go-to source for reliable answers. Come back soon for more expert insights.
