Answer :
Let's approach this step-by-step.
### Given Information:
- The 5th term in a geometric series (GS) is 81.
- The 8th term in the same GS is 2187.
### Step 1: Understand the Geometric Series
A geometric series has terms of the form [tex]\(a, ar, ar^2, ar^3, \ldots\)[/tex] where:
- [tex]\(a\)[/tex] is the first term.
- [tex]\(r\)[/tex] is the common ratio.
### Step 2: Formulate Equations for the Given Series
From the given information:
1. The 5th term is [tex]\(ar^4 = 81\)[/tex]
2. The 8th term is [tex]\(ar^7 = 2187\)[/tex]
### Step 3: Solve for the Common Ratio and First Term
Set up the equations:
[tex]\[ a r^4 = 81 \quad \text{(1)} \][/tex]
[tex]\[ a r^7 = 2187 \quad \text{(2)} \][/tex]
To solve for [tex]\(r\)[/tex], divide equation (2) by equation (1):
[tex]\[ \frac{a r^7}{a r^4} = \frac{2187}{81} \][/tex]
[tex]\[ r^3 = 27 \][/tex]
[tex]\[ r = 3 \][/tex]
Now substitute [tex]\(r = 3\)[/tex] back into equation (1) to find [tex]\(a\)[/tex]:
[tex]\[ a \cdot 3^4 = 81 \][/tex]
[tex]\[ a \cdot 81 = 81 \][/tex]
[tex]\[ a = 1 \][/tex]
So, the first term [tex]\(a\)[/tex] is 1, and the common ratio [tex]\(r\)[/tex] is 3.
### Step 4: Find [tex]\(k\)[/tex] for the Given Conditions
#### (a) [tex]\(k, k+2,\)[/tex] and [tex]\(k+6\)[/tex] are in GS:
For the terms [tex]\(k, k+2, k+6\)[/tex] to be in GS, the ratio of consecutive terms must be equal:
[tex]\[ \frac{k+2}{k} = \frac{k+6}{k+2} \][/tex]
Cross-multiplying gives:
[tex]\[ (k+2)^2 = k(k+6) \][/tex]
Expanding both sides:
[tex]\[ k^2 + 4k + 4 = k^2 + 6k \][/tex]
Simplifying, we get:
[tex]\[ 4 = 2k \][/tex]
[tex]\[ k = 2 \][/tex]
So, [tex]\(k = 2\)[/tex].
#### (b) [tex]\(k+1, 2k,\)[/tex] and [tex]\(k+6\)[/tex] are in GS:
For the terms [tex]\(k+1, 2k, k+6\)[/tex] to be in GS, the ratio of consecutive terms must be equal:
[tex]\[ \frac{2k}{k+1} = \frac{k+6}{2k} \][/tex]
Cross-multiplying gives:
[tex]\[ 2k \cdot 2k = (k+1)(k+6) \][/tex]
[tex]\[ 4k^2 = k^2 + 7k + 6 \][/tex]
Rearrange to form a quadratic equation:
[tex]\[ 4k^2 - k^2 - 7k - 6 = 0 \][/tex]
[tex]\[ 3k^2 - 7k - 6 = 0 \][/tex]
Solve this quadratic equation using the quadratic formula, [tex]\(k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex]:
[tex]\[ k = \frac{7 \pm \sqrt{49 + 72}}{6} \][/tex]
[tex]\[ k = \frac{7 \pm \sqrt{121}}{6} \][/tex]
[tex]\[ k = \frac{7 \pm 11}{6} \][/tex]
So, we get two solutions:
[tex]\[ k = \frac{18}{6} = 3 \][/tex]
[tex]\[ k = \frac{-4}{6} = -\frac{2}{3} \][/tex]
Therefore, the possible values of [tex]\(k\)[/tex] are [tex]\(-\frac{2}{3}\)[/tex] and [tex]\(3\)[/tex].
### Final Results
- For (a), the value of [tex]\(k\)[/tex] is [tex]\(2\)[/tex].
- For (b), the values of [tex]\(k\)[/tex] are [tex]\(-\frac{2}{3}\)[/tex] and [tex]\(3\)[/tex].
### Given Information:
- The 5th term in a geometric series (GS) is 81.
- The 8th term in the same GS is 2187.
### Step 1: Understand the Geometric Series
A geometric series has terms of the form [tex]\(a, ar, ar^2, ar^3, \ldots\)[/tex] where:
- [tex]\(a\)[/tex] is the first term.
- [tex]\(r\)[/tex] is the common ratio.
### Step 2: Formulate Equations for the Given Series
From the given information:
1. The 5th term is [tex]\(ar^4 = 81\)[/tex]
2. The 8th term is [tex]\(ar^7 = 2187\)[/tex]
### Step 3: Solve for the Common Ratio and First Term
Set up the equations:
[tex]\[ a r^4 = 81 \quad \text{(1)} \][/tex]
[tex]\[ a r^7 = 2187 \quad \text{(2)} \][/tex]
To solve for [tex]\(r\)[/tex], divide equation (2) by equation (1):
[tex]\[ \frac{a r^7}{a r^4} = \frac{2187}{81} \][/tex]
[tex]\[ r^3 = 27 \][/tex]
[tex]\[ r = 3 \][/tex]
Now substitute [tex]\(r = 3\)[/tex] back into equation (1) to find [tex]\(a\)[/tex]:
[tex]\[ a \cdot 3^4 = 81 \][/tex]
[tex]\[ a \cdot 81 = 81 \][/tex]
[tex]\[ a = 1 \][/tex]
So, the first term [tex]\(a\)[/tex] is 1, and the common ratio [tex]\(r\)[/tex] is 3.
### Step 4: Find [tex]\(k\)[/tex] for the Given Conditions
#### (a) [tex]\(k, k+2,\)[/tex] and [tex]\(k+6\)[/tex] are in GS:
For the terms [tex]\(k, k+2, k+6\)[/tex] to be in GS, the ratio of consecutive terms must be equal:
[tex]\[ \frac{k+2}{k} = \frac{k+6}{k+2} \][/tex]
Cross-multiplying gives:
[tex]\[ (k+2)^2 = k(k+6) \][/tex]
Expanding both sides:
[tex]\[ k^2 + 4k + 4 = k^2 + 6k \][/tex]
Simplifying, we get:
[tex]\[ 4 = 2k \][/tex]
[tex]\[ k = 2 \][/tex]
So, [tex]\(k = 2\)[/tex].
#### (b) [tex]\(k+1, 2k,\)[/tex] and [tex]\(k+6\)[/tex] are in GS:
For the terms [tex]\(k+1, 2k, k+6\)[/tex] to be in GS, the ratio of consecutive terms must be equal:
[tex]\[ \frac{2k}{k+1} = \frac{k+6}{2k} \][/tex]
Cross-multiplying gives:
[tex]\[ 2k \cdot 2k = (k+1)(k+6) \][/tex]
[tex]\[ 4k^2 = k^2 + 7k + 6 \][/tex]
Rearrange to form a quadratic equation:
[tex]\[ 4k^2 - k^2 - 7k - 6 = 0 \][/tex]
[tex]\[ 3k^2 - 7k - 6 = 0 \][/tex]
Solve this quadratic equation using the quadratic formula, [tex]\(k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex]:
[tex]\[ k = \frac{7 \pm \sqrt{49 + 72}}{6} \][/tex]
[tex]\[ k = \frac{7 \pm \sqrt{121}}{6} \][/tex]
[tex]\[ k = \frac{7 \pm 11}{6} \][/tex]
So, we get two solutions:
[tex]\[ k = \frac{18}{6} = 3 \][/tex]
[tex]\[ k = \frac{-4}{6} = -\frac{2}{3} \][/tex]
Therefore, the possible values of [tex]\(k\)[/tex] are [tex]\(-\frac{2}{3}\)[/tex] and [tex]\(3\)[/tex].
### Final Results
- For (a), the value of [tex]\(k\)[/tex] is [tex]\(2\)[/tex].
- For (b), the values of [tex]\(k\)[/tex] are [tex]\(-\frac{2}{3}\)[/tex] and [tex]\(3\)[/tex].
