Step-by-step explanation:
1. In an arithmetic sequence, the general term can be written as
xₙ = y + d(a-1), where xₐ represents the ath term, y is the first value, and d is the common difference.
Given the third term and the fifth term, and knowing that the difference between each term is d, we can say that the 4th term is x₃+d and the fifth term is the fourth term plus d, or (x₃+d)+d =
x₃+2d. =x₅ Given x₃ and x₅, we can subtract x₃ from both sides to get
x₅-x₃ = 2d
divide by 2 to isolate d
(x₅-x₃)/2 = d
This lets us solve for d. Given d, we can say that
x₃ = y+d(2)
subtract 2*d from both sides to isolate the y
x₃ -2*d = y
Therefore, because we know x₃ and d at this point, we can solve for y, letting us plug y and d into our original equation of
xₙ = y + d(a-1)
2.
Given the third and fifth term, with a common ratio of r, we can say that the fourth term is x₃ * r. Then, the fifth term is
x₃* r * r
= x₃*r² = x₅
divide both sides by x₃ to isolate the r²
x₅/x₃ = r²
square root both sides
√(x₅/x₃) = ±r
One thing that is important to note is that we don't know whether r is positive or negative. For example, if x₃ = 4 and x₅ = 16, regardless of whether r is equal to 2 or -2, 4*r² = 16. I will be assuming that r is positive for this question.
Given the common ratio, we can find x₆ as x₅ * r, x₇ as x₅*r², and all the way up to x₁₀ = x₅*r⁵. We don't know the general term, but can still find the tenth term of the sequence
<span>Θ = area of a sector
