Question

NO LINKS!! Please help me with these sequences Part 2x​

Answer 1

4. 121,800

5. 591 3/32

6. 1/4

Explanation:

4. Sum of arithmetic sequence

You want the sum of the first 200 terms of the arithmetic sequence starting 12, 18, 24, ....

The sum of the first n terms of an arithmetic sequence is given by the formula ...

Sn = (2·a1 +d(n -1))(n/2)

where a1 is the first term and d is the common difference. Your sequence has first term 12 and common difference 18-12 = 6. The desired sum is ...

S200 = (2·12 +6(200 -1))/(200/2) = (24 +1194)(100) = 121,800

5. Sum of geometric sequence

You want the sum of the first 8 terms of the geometric sequence starting 12, 18, 27, ....

The sum of the first n terms of a geometric sequence is given by the formula ...

Sn = a1·(r^n -1)/(r -1)

where a1 is the first term and r is the common ratio. Your sequence has first term 12 and common ratio 18/12 = 3/2. The desired sum is ...

S8 = 12·((3/2)^8 -1)/(3/2 -1) = 24·6305/256 = 591 3/32

6. Sum of geometric sequence

For this sequence, a1 = 1/12 and r = 2/3. When the sum is infinite and |r| < 1, the sum formula becomes ...

S = a1/(1 -r)

The desired sum is ...

S = (1/12)/(1 -2/3) = (1/12)/(4/12) = 1/4

Answer 2

`\textsf{4.} \quad 121,800`

`\textsf{5.} \quad (18915)/(32)=591.09375`

`\textsf{6.} \quad (1)/(4)`

Explanation:

Question 4

`\boxed{\begin{minipage}{7.3 cm}\underline{Sum of the first $n$ terms of an arithmetic series}\\\\$S_n=(1)/(2)n[2a+(n-1)d]$\\\\where:\\\phantom{ww}$\bullet$ $a$ is the first term. \\ \phantom{ww}$\bullet$ $d$ is the common difference.\\ \phantom{ww}$\bullet$ $n$ is the position of the term.\\\end{minipage}}`

Given arithmetic sequence:

• 12, 18, 24, ...

The first term of the given sequence is 12.

The common difference can be found by subtracting the first term from the second term:

`\implies d=a_2-a_1=18-12=6`

Therefore:

• a = 12
• d = 6

To find the sum of the first 200 terms, substitute n = 200, a = 12 and d = 6 into the formula:

`\begin{aligned}S_(200)&=(1)/(2)(200)[2(12)+(200-1)(6)]\\&=100[24+(199)(6)]\\&=100[24+1194]\\&=100[1218]\\&=121800\end{aligned}`

Question 5

`\boxed{\begin{minipage}{7.3 cm}\underline{Sum of the first $n$ terms of a geometric series}\\\\$S_n=(a(1-r^n))/(1-r)$\\\\where:\\\phantom{ww}$\bullet$ $a$ is the first term. \\ \phantom{ww}$\bullet$ $r$ is the common ratio.\\ \phantom{ww}$\bullet$ $n$ is the position of the term.\\\end{minipage}}`

Given geometric sequence:

• 12, 18, 27, ...

The first term of the given sequence is 12.

The common ratio can be found by dividing the second term by the first term:

`\implies r=(a_2)/(a_1)=(18)/(12)=1.5`

Therefore:

• a = 12
• r = 1.5

To find the sum of the first 8 terms, substitute n = 8, a = 12 and r = 1.5 into the formula:

`\begin{aligned}\implies S_(8)&=(12(1-1.5^8))/(1-1.5)\\\\&=(12\left(1-(6561)/(256)\right))/(-0.5)\\\\&=(12\left(-(6305)/(256)\right))/(-0.5)\\\\&=(-(18915)/(64))/(-0.5)\\\\&=(18915)/(32)\\\\&=591.09375\end{aligned}`

Question 6

`\boxed{\begin{minipage}{5.5 cm}\underline{Sum of an infinite geometric series}\\\\$S_(\infty)=(a)/(1-r)$\\\\where:\\\phantom{ww}$\bullet$ $a$ is the first term. \\ \phantom{ww}$\bullet$ $r$ is the common ratio.\\\end{minipage}}`

Given geometric sequence:

• 
  `(1)/(12),(1)/(18),(1)/(27),...`

The first term of the given sequence is ¹/₁₂.

The common ratio can be found by dividing the second term by the first term:

`\implies r=(a_2)/(a_1)=((1)/(18))/((1)/(12))=(2)/(3)`

Therefore:

• a = ¹/₁₂
• r = ²/₃

To find the sum of the infinite geometric sequence, substitute a = ¹/₁₂ and r = ²/₃ into the formula:

`\begin{aligned}\implies S_(\infty)&=((1)/(12))/(1-(2)/(3))\\\\&=((1)/(12))/((1)/(3))\\\\&=(1)/(12) * (3)/(1)\\\\&=(3)/(12)\\\\&=(1)/(4)\end{aligned}`