What is the recursive formula for this geometric sequence? -4,-24,-144,-864,...

1 answer:

7 0

$\bf -4~~,~~\stackrel{-4\cdot 6}{-24}~~,~~\stackrel{-24\cdot 6}{-144}~~,~~\stackrel{-144\cdot 6}{-864}~\hfill \impliedby \begin{array}{llll} \textit{we're multiplying by 6 the previous}\\ \textit{term(n-1) to get the current one} \end{array} \\\\[-0.35em] ~\dotfill\\\\ a_n = a_{n-1}(6)\qquad for~~a_1=-4~~,~~n>2\qquad \impliedby \textit{recursive formula}$

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olga2289 [7]

$\text{The sum of } 1\frac{7}{10} \text{ and } 3\frac{3}{5} \text{ is } 5\frac{3}{10}$

Solution:

Given that we have to find the sum of $1\frac{7}{10} \text{ and } 3\frac{3}{5}$

Let us first convert the mixed fractions to improper fractions

Multiply the whole number part by the fraction's denominator.

Add that to the numerator.

Then write the result on top of the denominator

Therefore,

$1\frac{7}{10} = \frac{10 \times 1 + 7}{10} = \frac{10+7}{10} = \frac{17}{10}\\\\3\frac{3}{5} = \frac{5 \times 3 + 3}{5} = \frac{18}{5}$

Now we can add both the fractions

$1\frac{7}{10} + 3\frac{3}{5} = \frac{17}{10} + \frac{18}{5}$

Make the denominators same

$1\frac{7}{10} + 3\frac{3}{5} = \frac{17}{10} + \frac{18 \times 2}{5 \times 2}\\\\1\frac{7}{10} + 3\frac{3}{5} = \frac{17}{10} + \frac{36}{10}\\\\\text{Add the fractions since the denominator is same }\\\\1\frac{7}{10} + 3\frac{3}{5} = \frac{17+36}{10} = \frac{53}{10}$