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3 years ago

13

Fill in the missing exponents in each box & show reasoning (5^?)^3=5^1

2 answers:

Nutka1998 [239]3 years ago

3 0

the missing exponent would be 2

sesenic [268]3 years ago

3 0

$\bf (5^?)^{3}=5^1\implies 5^{?\cdot 3}=5^1\implies \stackrel{\textit{same base, the exponents must also be the same}}{?3=1\implies ?=\cfrac{1}{3}}$

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The first term of a geometric sequence is 15, and the 5th term of the sequence is <img src="https://tex.z-dn.net/?f=%5Cfrac%7B24

sladkih [1.3K]

The geometric sequence is $15,9,\frac{27}{5},\frac{81}{25}, \frac{243}{125}$

Explanation:

Given that the first term of the geometric sequence is 15

The fifth term of the sequence is $\frac{243}{125}$

We need to find the 2nd, 3rd and 4th term of the geometric sequence.

To find these terms, we need to know the common difference.

The common difference can be determined using the formula,

$a_n=a_1(r)^{n-1}$

where $a_1=15$ and $a_5=\frac{243}{125}$

For $n=5$ , we have,

$\frac{243}{125}=15(r)^4$

Simplifying, we have,

$r=\frac{3}{5}$

Thus, the common difference is $r=\frac{3}{5}$

Now, we shall find the 2nd, 3rd and 4th terms by substituting $n=2,3,4$ in the formula $a_n=a_1(r)^{n-1}$

For $n=2$

$a_2=15(\frac{3}{5} )^{1}$

$=9$

Thus, the 2nd term of the sequence is 9

For $n=3$ , we have,

$a_3=15(\frac{3}{5} )^{2}$

$=15(\frac{9}{25} )$

$=\frac{27}{5}$

Thus, the 3rd term of the sequence is $\frac{27}{5}$

For $n=4$ , we have,

$a_4=15(\frac{3}{5} )^{3}$

$=15(\frac{27}{25} )$

$=\frac{81}{25}$

Thus, the 4th term of the sequence is $\frac{81}{25}$

Therefore, the geometric sequence is $15,9,\frac{27}{5},\frac{81}{25}, \frac{243}{125}$

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3 years ago