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sergij07 [2.7K]

3 years ago

14

Shhhh Help me and you get Star ratings

2 answers:

san4es73 [151]3 years ago

4 0

Answer: find the measure of <6

Step-by-step explanation:

lidiya [134]3 years ago

4 0

Answer:

Interior angles = 2, 4, 5, 7

Exterior angles = 1, 3, 6, 8

Alternate interior angles = 2 & 7, 5 & 4

Alternate exterior angles = 1 & 8, 6 & 3

<3 = 96*

<5 = 84*

<6 = 96*

Step-by-step explanation:

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What is the value of x? <br> 4 <br> 8<br> 6<br> 10

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The right answer for the question is 6.

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What is the degree of the following polynomial: . 3x3 + 2x - 4

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The sum of the first n terms of a geometric series is 364? The sum of their reciprocals 364/243. If the first term is 1, find n

Afina-wow [57]

If the geometric series has first term $a$ and common ratio $r$ , then its $N$ -th partial sum is

$\displaystyle S_N = \sum_{n=1}^N ar^{n-1} = a + ar + ar^2 + \cdots + ar^{N-1}$

Multiply both sides by $r$ , then subtract $rS_N$ from $S_N$ to eliminate all the middle terms and solve for $S_N$ :

$rS_N = ar + ar^2 + ar^3 + \cdots + ar^N$

$\implies (1 - r) S_N = a - ar^N$

$\implies S_N = \dfrac{a(1-r^N)}{1-r}$

The $N$ -th partial sum for the series of reciprocal terms (denoted by $S'_N$ ) can be computed similarly:

$\displaystyle S'_N = \sum_{n=1}^N \frac1{ar^{N-1}} = \frac1a + \frac1{ar} + \frac1{ar^2} + \cdots + \frac1{ar^{N-1}}$

$\dfrac{S'_N}r = \dfrac1{ar} + \dfrac1{ar^2} + \dfrac1{ar^3} + \cdots + \dfrac1{ar^N}$

$\implies \left(1 - \dfrac1r\right) S'_N = \dfrac1a - \dfrac1{ar^N}$

$\implies S'_N = \dfrac{1 - \frac1{r^N}}{a\left(1 - \frac1r\right)} = \dfrac{r^N - 1}{a(r^N - r^{N-1})} = \dfrac{1 - r^N}{a r^{N-1} (1 - r)}$

We're given that $a=1$ , and the sum of the first $n$ terms of the series is

$S_n = \dfrac{1-r^n}{1-r} = 364$

and the sum of their reciprocals is

$S'_n = \dfrac{1 - r^n}{r^{n-1}(1 - r)} = \dfrac{364}{243}$

By substitution,

$\dfrac{1 - r^n}{r^{n-1}(1-r)} = \dfrac{364}{r^{n-1}} = \dfrac{364}{243} \implies r^{n-1} = 243$

Manipulating the $S_n$ equation gives

$\dfrac{1 - r^n}{1-r} = 364 \implies r (364 - r^{n-1}) = 363$

so that substituting again yields

$r (364 - 243) = 363 \implies 121r = 363 \implies \boxed{r=3}$

and it follows that

$r^{n-1} = 243 \implies 3^{n-1} = 3^5 \implies n-1 = 5 \implies \boxed{n=6}$

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