7x + 12 = 6 solve for x

Given the sequence in the table below, determine the sigma notation of the sum for term 4 through term 15. N an 1 4 2 −12 3 36 t

Rzqust [24]

By applying basic property of Geometric progression we can say that sum of 15 terms of a sequence whose first three terms are 5, -10 and 2 is $\sum_{n=4}^{15} 5(-2)^{n-1}$$$

<h3>What is sequence ?</h3>

Sequence is collection of numbers with some pattern .

Given sequence

$a_{1}=5\\\\a_{2}=-10\\\\\\a_{3}=20$

We can see that

$\frac{a_1}{a_2}=\frac{-10}{5}=-2\\$

and

$\frac{a_2}{a_3}=\frac{20}{-10}=-2\\$

Hence we can say that given sequence is Geometric progression whose first term is 5 and common ratio is -2

Now $n^{th}$ term of this Geometric progression can be written as

$T_{n}= 5\times(-2)^{n-1}$

So summation of 15 terms can be written as

$\sum_{n=4}^{15} T_{n}\\\\$\\$\sum_{n=4}^{15} 5(-2)^{n-1}$$$

By applying basic property of Geometric progression we can say that sum of 15 terms of a sequence whose first three terms are 5, -10 and 2 is $\sum_{n=4}^{15} 5(-2)^{n-1}$$$

To learn more about Geometric progression visit : brainly.com/question/14320920

8 0

2 years ago

A sequence of two transformations that can be used to show that polygon ABCDE is congruent to polygon GFJIH

Burka [1]

Two transformations that can be used to show that polygon ABCDE is congruent to polygon GFJIH : <u>rotation -90° and translation (0,2)</u>

<h3>Further explanation</h3>

There are several forms of transformation, including translation, dilation, rotation or reflection

There are 2 shapes of geometry:

Similar figures:

the same shape but different in size,

Congruent figures:

the same size and the same shape

Both polygons will appear congruent if, after being transformed, the two polygons will map exactly to each other

A sequence of two transformations e can do is :

1. rotation -90°

Let point H(4,1) from polygon GFJIH , we rotate -90°

$\left[\begin{array}{ccc}x'\\y'\end{array}\right]=\left[\begin{array}{ccc}0&1\\-1&0\end{array}\right]\left[\begin{array}{ccc}4-6\\1+1\end{array}\right]+\left[\begin{array}{ccc}6\\-1\end{array}\right]\\\\=\left[\begin{array}{ccc}0&1\\-1&0\end{array}\right]\left[\begin{array}{ccc}-2\\2\end{array}\right]+\left[\begin{array}{ccc}6\\-1\end{array}\right]\\\\=\left[\begin{array}{ccc}8\\1\end{array}\right]$