Two questions I need help with.
Find the indicated partial sum using the information given.
1. a3=4/9, a7=9/64, S6
Determine wheater the infinite geometric series has a finite sum
2. 6+ 3+ 3/2 + 3/4 +....
3. -49 + (-7) + (- 1/7) +...
1. S6 = 6/2(a+5d)
d = (A7-A3)/4
a = A3-2d
2. since |r| = 1/2 < 1, yes the sum is finits
3. same
I will assume that by a3 you mean term(3)
and the terms form an geometric sequence
term(3) = ar^2 = 4/9
term(7) = ar^6 = 9/64
divide term7 by term3
r^4 = (9/64) / (4/9) = 81/256
r = ± 3/4
the in ar^2 = 4/9
a(9/16) = 4/9
a = 64/81
so Sum(6) = a(r^6 - 1)/(r-1)
= (64/81)((3/4)^6 - 1)/(3/4 - 1)
= 3367/1296
= appr 2.59799
2. term2/term1 = 1/2
term3/term2 = 1/2
since |r| < 1 , the infinite series has a finite sum
2nd one:
r = -7/-49 = 1/7
yes, so does that one
Sure! I can help you with both questions.
For the first question,
1. a₃ = 4/9 and a₇ = 9/64, and you need to find the sum of the first 6 terms, represented as S₆.
To find the partial sum, you can use the geometric series formula:
Sₙ = a₁ * (1 - rⁿ) / (1 - r)
In this formula, a₁ is the first term, r is the common ratio, and n is the total number of terms. However, we only have the values of a₃ and a₇, so we need to find the common ratio (r) first.
To find the common ratio (r), we can divide any term by the previous term:
r = a₇ / a₃ = (9/64) / (4/9)
Simplifying this expression:
r = (9/64) * (9/4) = 81/256
Now that we have the common ratio (r), we can substitute the given values into the formula, using n = 6:
S₆ = a₁ * (1 - r⁶) / (1 - r)
But we don't know a₁, the first term. To find it, we can use the formula:
a₇ = a₁ * r⁴
Substituting the values:
(9/64) = a₁ * (81/256)⁴
Simplifying this expression:
(9/64) = a₁ * 81⁴ / 256⁴
(9/64) = a₁ * 531441 / 4294967296
(9/64) * 4294967296 = a₁ * 531441
a₁ = [(9/64) * 4294967296] / 531441
Finally, substitute the found value of a₁ into the original formula to find S₆.
For the second question,
2. The series 6 + 3 + 3/2 + 3/4 + ... is an infinite geometric series.
To determine if this series has a finite sum, we need to check if the absolute value of the common ratio (r) is less than 1.
The common ratio (r) can be found by dividing any term by the previous term:
r = (3/2) / 3 = 1/2
Since the absolute value of the common ratio (r = 1/2) is less than 1, this series does have a finite sum.
For the third question,
3. The series -49 + (-7) + (-1/7) + ... is an infinite geometric series.
Similar to the second question, we need to check if the absolute value of the common ratio (r) is less than 1 to determine if the series has a finite sum.
The common ratio (r) can be found by dividing any term by the previous term:
r = (-1/7) / (-7) = 1/49
Since the absolute value of the common ratio (r = 1/49) is less than 1, this series also has a finite sum.