Answer:
Q7. 945; Q8. 45 927; Q9. 546.75
Step-by-step explanation:
7. Arithmetic sequence
The explicit formula for the nth term of an arithmetic sequence is
aₙ = a₁ + d(n - 1 )
a₁ is the first term, and d is the difference in value between consecutive terms. Thus,
a₁ = 91, d = -4, n = 15
a₁₅ = 91 - 4(15 - 1) = 91 - 4(14) = 91 - 56 = 35
We find the sum of an arithmetic series by multiplying the number of terms by the average of the first and last terms.
S = n[(a₁ + a₁₅)/2]
S = 15[(91 + 35)/2] = 15 × 126/2 = 945
The sum of the first 15 terms is 945.
8. nth term of geometric sequence
a₃ = 63 and r = -3
The explicit formula for the nth term of a geometric sequence is
aₙ = a₁rⁿ⁻¹
a₁ is the first term and r is the common ratio.
If we start counting from a₃, then a₉ is the seventh term in the sequence.
In your sequence, r = -3.
a₇ = 63(-3)⁷⁻¹ = 63(-3)⁶ = 63 × 729 = 45 927
The ninth term in the sequence is 45 927.
9. Sum of geometric series
a₁ = 729, aₙ = -3, r = -⅓
I don't know what you mean by aₙ = -3. It says that every term is -3, so I am going to ignore it.
Since |r| <1, we have a convergent series, and the formula for the sum is
S = a₁/(1 - r)
∴ S = 729/[(1 - (-⅓)] = 729/1⅓ = 729/(⁴/₃) = 729 × ¾ = 546.75
The sum of the geometric series is 546.75.
7 Answer: 1060
Step-by-step explanation:
[tex]a_1=91\qquad d=-4\qquad a_n=15\\\\\text{First, find n:}\\a_n=a_1+d(n-1)\\\\15=91-4(n-1)\\-76 =-4(n-1)\\19=n-1\\20=n\\\\\text{Now find the sum:}\\S_n=\dfrac{a_1+a_n}{2}\cdot n\\\\\\S_{20}=\dfrac{91+15}{2}\cdot 20\\\\\\.\quad =\dfrac{106}{2}\cdot 20\\\\\\.\quad =106\cdot 10\\\\.\quad =\large\boxed{1060}[/tex]
8 Answer: 45,927
Step-by-step explanation:
[tex]\text{Find the first term }(a_1):\\a_3=63\qquad r=-3\qquad n=3\\\\a_n=a_1\cdot r^{n-1}\\63=a_1\cdot (-3)^{3-1}\\63=a_1\cdot (-3)^2\\63=a_1\cdot 9\\7=a_1\\\\\\\text{Now find the ninth term:}\\a_1=7\qquad r=-3\qquad n=9\\\\a_n=a_1\cdot r^{n-1}\\a_9=7\cdot (-3)^{9-1}\\.\quad =7\cdot (-3)^{8}\\.\quad =7\cdot 6561\\.\quad =\large\boxed{45,927}[/tex]
9 Answer: 546
Step-by-step explanation:
[tex]\text{Find n:}\\a_1=729\qquad a_n=-3\qquad r=-\dfrac{1}{3}\\\\a_n=a_1\cdot r^{n-1}\\-3=729\cdot \bigg(-\dfrac{1}{3}\bigg)^{n-1}\\\\-\dfrac{3}{729}=\bigg(-\dfrac{1}{3}\bigg)^{n-1}\\\\-\dfrac{1}{243}=\bigg(-\dfrac{1}{3}\bigg)^{n-1}\\\\\bigg(-\dfrac{1}{3}\bigg)^5=\bigg(-\dfrac{1}{3}\bigg)^{n-1}\\\\5=n-1\\6=n\\\\\\\text{Find the sum:}\\a_1=729\qquad r=-\dfrac{1}{3}\quad n=6\\\\S_n=\dfrac{a_1(1-r^n)}{1-r}[/tex]
[tex]S_6=\dfrac{729(1-(-\frac{1}{3})^6)}{1-(-\frac{1}{3})}\\\\\\.\quad =\dfrac{729(\frac{728}{729})}{\frac{4}{3}}\\\\\\.\quad =728\cdot \dfrac{3}{4}\\\\.\quad =128\cdot 3\\\\.\quad = \large\boxed{546}[/tex]
