Question

If the first term of a positive geometric series is 3 and 3rd term is 12, find the sum of its first 8 terms.​

Answer 1

To find the sum of the first 8 terms of a positive geometric series, given that the first term is 3 and the third term is 12, we can follow these steps: 1. In a geometric series, each term is found by multiplying the previous term by a constant factor called the common ratio. 2. Let's assume the common ratio of the geometric series is represented by the variable "r". 3. The first term of the series is given as 3, so we have the equation: a₁ = 3. 4. The third term of the series is given as 12, so we can express it in terms of the first term and the common ratio: a₃ = a₁ × r² = 3 × r² = 12. 5. Simplifying the equation, we have r² = 12/3 = 4. 6. Taking the square root of both sides, we find r = ±2. However, since the series is defined as positive, we take the positive value: r = 2. 7. Now that we know the common ratio, we can find the sum of the first 8 terms using the formula for the sum of a geometric series: - The sum of the first 8 terms, denoted as S₈, can be calculated using the formula S₈ = a₁ × (1 - r⁸) / (1 - r). - Plugging in the values we have, S₈ = 3 × (1 - 2⁸) / (1 - 2). - Simplifying further, S₈ = 3 × (1 - 256) / (1 - 2). - S₈ = 3 × (-255) / (-1) = 765. 8. Therefore, the sum of the first 8 terms of the given geometric series is 765. I hope this explanation helps you understand how to find the sum of the first 8 terms of a geometric series. If you have any further questions, feel free to ask.

Explanation:

Answer 2

Sum of first 8th terms is 765.

Explanation:

A geometric series is a series of terms where each term is multiplied by a constant ratio to get the next term. The constant ratio is called the common ratio.

The sum of a geometric series can be calculated using the following formula:

`\boxed{\sf S_n =( a(1 - r^n))/((1 - r))}`

and it's nth term is calculated by using formula:

`\boxed{\sf t_n=ar^(n-1)}`

where:

• Sn is the sum of the first n terms.
• tn is nth term of the series
• a is the first term.
• r is the common ratio
• n is the number of terms

For the Question:

`\textsf{ 1st term(a) = 3}`

`\textsf{3rd term}\sf (t_3)= 12`

By using nth term formula, we can find common ratio(r).

`\sf t_3 =ar^(n-1)`

substituting value

`\sf t_3 =3r^(3-1)`

`\sf 12 =3r^2`

dividing both side by 3, we get

`\sf r^2 =(12)/(3)`

`\sf r^2 = 4`

square rooting on both side

`\sf r = √(4)`

`\sf r = 2`

Therefore, common ratio is 2.

Again,

Let's find the sum of 8th terms by using formula:

`\boxed{\sf S_n =( a(1 - r^n))/((1 - r))}`

In this case,

• a = 3
• r = 2
• n =8

Substituting value,

`\sf S_8 = (3(1-2^8))/((1-2))`

`\sf S_8 = (3(1-256))/(-1)`

`\sf S_8 = (3(-255))/(-1)`

`\sf S_8 = (-765)/(-1)`

`\sf S_8 = 765`

Therefore, Sum of first 8th terms is 765.