Question

Show work and explain with formulas.

7. Find the sum: a_1 = 91, d = -4, a_n = 15

8. Find the ninth term of a geometric sequence in which a_3 = 63 and r = -3

9. Find the sum of geometric series for which a_1 = 729, a_n = -3, r = -1/3​

Answer 1

7 Answer: 1060

Explanation:

`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=(a_1+a_n)/(2)\cdot n\\\\\\S_(20)=(91+15)/(2)\cdot 20\\\\\\.\quad =(106)/(2)\cdot 20\\\\\\.\quad =106\cdot 10\\\\.\quad =\large\boxed{1060}`

8 Answer: 45,927

Explanation:

`\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}`

9 Answer: 546

Explanation:

`\text{Find n:}\\a_1=729\qquad a_n=-3\qquad r=-(1)/(3)\\\\a_n=a_1\cdot r^(n-1)\\-3=729\cdot \bigg(-(1)/(3)\bigg)^(n-1)\\\\-(3)/(729)=\bigg(-(1)/(3)\bigg)^(n-1)\\\\-(1)/(243)=\bigg(-(1)/(3)\bigg)^(n-1)\\\\\bigg(-(1)/(3)\bigg)^5=\bigg(-(1)/(3)\bigg)^(n-1)\\\\5=n-1\\6=n\\\\\\\text{Find the sum:}\\a_1=729\qquad r=-(1)/(3)\quad n=6\\\\S_n=(a_1(1-r^n))/(1-r)`

`S_6=(729(1-(-(1)/(3))^6))/(1-(-(1)/(3)))\\\\\\.\quad =(729((728)/(729)))/((4)/(3))\\\\\\.\quad =728\cdot (3)/(4)\\\\.\quad =128\cdot 3\\\\.\quad = \large\boxed{546}`

Answer 2

Q7. 945; Q8. 45 927; Q9. 546.75

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.