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san4es73 [151]
3 years ago
5

Find two consecutive integers such that the smaller integer is six more than twice the larger integer

Mathematics
1 answer:
ZanzabumX [31]3 years ago
8 0
Let the integers be x and x+1.

x = 6 + 2(x + 1)

x = 6 + 2x + 2

x = -8

Hence, the integers are -8 and -7.
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\bf \stackrel{\textit{first year}}{2000}~~,~~\stackrel{\textit{second year}}{2000+\stackrel{\textit{10\% of 2000}}{\frac{2000}{10}}}\implies 2200

now, if we take 2000 to be the 100%, what is 2200? well, 2200 is just 100% + 10%, namely 110%, and if we change that percent format to a decimal, we simply divide it by 100, thus \bf 110\%\implies \cfrac{110}{100}\implies 1.1.

so, 1.1 is the decimal number we multiply a term to get the next term, namely 1.1 is the common ratio.

\bf \qquad \qquad \textit{sum of a finite geometric sequence}\\\\S_n=\sum\limits_{i=1}^{n}\ a_1\cdot r^{i-1}\implies S_n=a_1\left( \cfrac{1-r^n}{1-r} \right)\quad \begin{cases}n=n^{th}\ term\\a_1=\textit{first term's value}\\r=\textit{common ratio}\\----------\\a_1=2000\\r=1.1\\n=4\end{cases}\\\\\\S_4=2000\left[ \cfrac{1-(1.1)^4}{1-1.1} \right]\implies S_4=2000\left(\cfrac{-0.4641}{-0.1}  \right)\\\\\\S_4=2000(4.641)\implies S_4=9282

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3 years ago
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