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The 1st term of a geometric sequence is 3 and the eighth term is 384. Find the common ratio, the sum and the product of the first 8 terms

User Teriiehina
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Answer:

  • ratio: 2
  • sum: 765
  • product: 1,761,205,026,816

Explanation:

The general term of a geometric sequence is ...

an = a1×r^(n-1)

Here, we're given that the first term is a1 = 3. From the 8th term, we can find the common ratio, r.

a8 = 384 = 3r^7

128 = r^7 . . . . . . . . divide by 3

2 = r . . . . . . . . . 7th root

The common ratio is 2.

__

The sum of the terms of a geometric series is ...

Sn = a1×(r^n -1)/(r -1)

For the known values of a1 and r, the sum o 8 terms is ...

S8 = 3×(2^8 -1)/(2 -1) = 3(255) = 765

The sum of the first 8 terms is 765.

__

The product of the first 8 terms will be ...

(3)(3×2)(3×2^2)(3×2^3)...(3×2^6)(3×2^7) = (3^8)(2^(1+2+3+...+7))

= (3^8)(2^28) = 6561 × 268,435,456

Many calculators cannot report the full value of this product. (Many spreadsheets can.)

The product of the first 8 terms is 1,761,205,026,816.

_____

In the attached, we found the value of r that made the 8th term be 384 by solving the equation 3×r^7 -384 = 0. The value is shown on the graph as the x-intercept of the exponential function. That x-intercept is 2, so r=2 (as above).

The 1st term of a geometric sequence is 3 and the eighth term is 384. Find the common-example-1
User Akshat Gupta
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