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3 years ago

The sum of two numbers is 61 . The larger number is 5 more than the smaller number. What are the numbers?

Mathematics

1 answer:

Elanso [62]3 years ago

6 0

Answer:

35.5 and 25.5

Step-by-step explanation:

all you have to do is take 61÷2=30.5 and you add five to one of the halves, and subtract five form the other.

61÷2= 30.5

30.5+5= 35.5

30.5-5= 25.5

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The first, third and thirteenth terms of an arithmetic sequence are the first 3 terms of a geometric sequence. If the first term

Salsk061 [2.6K]

Answer:

The first three terms of the geometry sequence would be $1$ , $5$ , and $25$ .

The sum of the first seven terms of the geometric sequence would be $127$ .

Step-by-step explanation:

Let $d$ denote the common difference of the arithmetic sequence.

Let $a_1$ denote the first term of the arithmetic sequence. The expression for the $n$ th term of this sequence (where $n\!$ is a positive whole number) would be $(a_1 + (n - 1)\, d)$ .

The question states that the first term of this arithmetic sequence is $a_1 = 1$ . Hence:

The third term of this arithmetic sequence would be $a_1 + (3 - 1)\, d = 1 + 2\, d$ .
The thirteenth term of would be $a_1 + (13 - 1)\, d = 1 + 12\, d$ .

The common ratio of a geometric sequence is ratio between consecutive terms of that sequence. Let $r$ denote the ratio of the geometric sequence in this question.

Ratio between the second term and the first term of the geometric sequence:

$\displaystyle r = \frac{1 + 2\, d}{1} = 1 + 2\, d$ .

Ratio between the third term and the second term of the geometric sequence:

$\displaystyle r = \frac{1 + 12\, d}{1 + 2\, d}$ .

Both $(1 + 2\, d)$ and $\left(\displaystyle \frac{1 + 12\, d}{1 + 2\, d}\right)$ are expressions for $r$ , the common ratio of this geometric sequence. Hence, equate these two expressions and solve for $d$ , the common difference of this arithmetic sequence.

$\displaystyle 1 + 2\, d = \frac{1 + 12\, d}{1 + 2\, d}$ .

$(1 + 2\, d)^{2} = 1 + 12\, d$ .

$d = 2$ .

Hence, the first term, the third term, and the thirteenth term of the arithmetic sequence would be $1$ , $(1 + (3 - 1) \times 2) = 5$ , and $(1 + (13 - 1) \times 2) = 25$ , respectively.

These three terms ( $1$ , $5$ , and $25$ , respectively) would correspond to the first three terms of the geometric sequence. Hence, the common ratio of this geometric sequence would be $r = 25 /5 = 5$ .

Let $a_1$ and $r$ denote the first term and the common ratio of a geometric sequence. The sum of the first $n$ terms would be:

$\displaystyle \frac{a_1 \, \left(1 - r^{n}\right)}{1 - r}$ .

For the geometric sequence in this question, $a_1 = 1$ and $r = 25 / 5 = 5$ .

Hence, the sum of the first $n = 7$ terms of this geometric sequence would be:

$\begin{aligned} & \frac{a_1 \, \left(1 - r^{n}\right)}{1 - r}\\ &= \frac{1 \times \left(1 - 2^{7}\right)}{1 - 2} \\ &= \frac{(1 - 128)}{(-1)} = 127 \end{aligned}$ .

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3 years ago

How do you solve a/6=15/2?

spayn [35]

15/2 = 7.5
7.5 times 6 = 45
Check work: 45/6 = 7.5

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