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cupoosta [38]
3 years ago
12

Find the indicated nth term show your solution 8,15,22,29,...,a18

Mathematics
1 answer:
Damm [24]3 years ago
4 0

Answer:

127

Step-by-step explanation:

Hey there! :)

Let us find the 18th term — first, there are two main types of sequences.

  • Arithmetic Sequence — a sequence that has same difference or common difference.
  • Geometric Sequence — a sequence that has same ratio.

But if you notice, it cannot be geometric sequence because if we do ratio test, there are no common ratios.

Hence, it can only be arithmetic sequence.

<u>C</u><u>o</u><u>m</u><u>m</u><u>o</u><u>n</u><u> </u><u>D</u><u>i</u><u>f</u><u>f</u><u>e</u><u>r</u><u>e</u><u>n</u><u>c</u><u>e</u>

\displaystyle \large{a_{n + 1} - a_n = d}

d is our common difference.

a_n+1 is the next term of a_n.

Let's check the common difference!

15-8 = 7

22-15 = 7

29-22 = 7

Therefore, 7 is our common difference.

<u>G</u><u>e</u><u>n</u><u>e</u><u>r</u><u>a</u><u>l</u><u> </u><u>T</u><u>e</u><u>r</u><u>m</u><u>s</u><u> </u><u>o</u><u>f</u><u> </u><u>A</u><u>r</u><u>i</u><u>t</u><u>h</u><u>m</u><u>e</u><u>t</u><u>i</u><u>c</u><u> </u><u>S</u><u>e</u><u>q</u><u>u</u><u>e</u><u>n</u><u>c</u><u>e</u>

\displaystyle \large{a_n = a_1  + (n - 1)d}

We know that a1 is the first term of sequence which is 8.

Our common difference is 7.

We want to find the 18th term — therefore,

\displaystyle \large{a_{18} = 8  + (18 - 1)7} \\  \displaystyle \large{a_{18} = 8  + (17)7} \\  \displaystyle \large{a_{18} = 8  + 119} \\  \displaystyle \large{a_{18} = 127}

Therefore, the 18th term is 127.

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From the question we can say that the Hexagon has three shapes inside it,

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Option D: Two irrational solutions

Explanation:

The equation is 17+3 x^{2}=6 x

Subtracting 6x from both sides, we have,

3x^{2} -6x+17=0

Solving the equation using quadratic formula,

x=\frac{6 \pm \sqrt{36-4(3)(17)}}{2(3)}

Simplifying the expression, we get,

\begin{aligned}x &=\frac{6 \pm \sqrt{36-204}}{6} \\&=\frac{6 \pm \sqrt{-168}}{6} \\&=\frac{6 \pm 2 i \sqrt{42}}{6}\end{aligned}

Taking out the common terms and simplifying, we have,

\begin{aligned}x &=\frac{2(3 \pm i \sqrt{42})}{6} \\&=\frac{(3 \pm i \sqrt{42})}{3}\end{aligned}

Dividing by 3, we get,

x=1+i \sqrt{\frac{14}{3}}, x=1-i \sqrt{\frac{14}{3}}

Hence, the equation has two irrational solutions.

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