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Find the number of terms in a GP given that its first and last terms are 16/3k and 243/ 256 k respectively and that its common ratio is 3/ 4 .

User Anamarie
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1 Answer

1 vote

Answer:

7

Explanation:

You want to know the number of terms in a geometric progression that has first term 16/3, last term 243/256, and a common ratio of 3/4.

Geometric Progression

The general term of a geometric progression is ...

an = a1(r^(n-1))

We want to find n when a1 = 16/3, an = 243/256, and r = 3/4.

Solution

243/256 = 16/3(3/4)^(n-1)

Multiplying by 3/16, we have ...

729/4096 = (3/4)^(n -1)

(3/4)^6 = (3/4)^(n-1) . . . . . . writing as powers of 3/4

6 = n -1 . . . . . . . . . . . . equating exponents

n = 7 . . . . . . . . . . add 1

The number of terms is 7.

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Additional comment

We can also use logarithms to find the value of n:

ln(729/4096) = (n -1)ln(3/4)

n = ln(729/4096)/ln(3/4) +1 = 7

The attachment shows this approach, along with the 7 terms of the sequence.

<95141404393>

Find the number of terms in a GP given that its first and last terms are 16/3k and-example-1
User Sovanlandy
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8.6k points