Find S^20 if the series 1 + 1.1 ..... is a (a) arithmetic (b) geometric

1 answer:

7 0

I'm guessing S^20 is actually referring to the 20th partial sum $S_{20}$ , with

$S_{20}=1+(1+1\times0.1)+(1+2\times0.1)+\cdots+(1+18\times0.1)+(1+19\times0.1)$

in the arithmetic case, and

$S_{20}=1+1.1+1.1^2+\cdots+1.1^{18}+1.1^{19}$

in the geometric case.

(a) Combining like terms, we have

$S_{20}=20+0.1(1+2+\cdots+18+19)$

and invoking the formula

$1+2+\cdots+(n-1)+n=\dfrac{n(n+1)}2$

we end up with

$S_{20}=20+0.1\dfrac{19\times20}2=39$

(b) Multiply $S_{20}$ by 1.1, then subtract this from $S_{20}$ :

$1.1S_{20}=1.1+1.1^2+1.1^3+\cdots+1.1^{19}+1.1^{20}$

$S_{20}-1.1S_{20}=-0.1S_{20}=1-1.1^{20}$
$\implies S_{20}=\dfrac{1.1^{20}-1}{0.1}\approx57.275$

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Answer :
x = 53.85

Step by step explanation:
If 78% of x (some numbers) is equal to 42, then we must set this as an equation to find it.

1) Rewriting 78% as a fraction and considering that when we say 78% of some number (x) we mean a product.
78% = 78/100 ➡️ 78/100x = 42

2) Then we can go on :
100 * 78/100x = 42 * 100
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3) Alternatively, simply dividing 42 by 0.78 (78%) would come to the same result :

42/0.78 = 53.85

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3 years ago

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Answer:

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Step-by-step explanation:

We have to perform an hypothesis test on the mean.

The null and alternative hypothesis are:

$H_0: \mu=10\\\\H_1: \mu \neq 10$