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How many terms of the arithmetic sequence {1,22,43,64,85,…} will give a sum of 2332? Show all steps including the formulas used

MA_775_DIABLO [31]

There's a slight problem with your question, but we'll get to that...

Consecutive terms of the sequence are separated by a fixed difference of 21 (22 = 1 + 21, 43 = 22 + 21, 64 = 43 + 21, and so on), so the n-th term of the sequence, a (n), is given recursively by

• a (1) = 1

• a (n) = a (n - 1) + 21 … … … for n > 1

We can find the explicit rule for the sequence by iterative substitution:

a (2) = a (1) + 21

a (3) = a (2) + 21 = (a (1) + 21) + 21 = a (1) + 2×21

a (4) = a (3) + 21 = (a (1) + 2×21) + 21 = a (1) + 3×21

and so on, with the general pattern

a (n) = a (1) + 21 (n - 1) = 21n - 20

Now, we're told that the sum of some number N of terms in this sequence is 2332. In other words, the N-th partial sum of the sequence is

a (1) + a (2) + a (3) + … + a (N - 1) + a (N) = 2332

or more compactly,

$\displaystyle\sum_{n=1}^N a(n) = 2332$

It's important to note that N must be some positive integer.

Replace a (n) by the explicit rule:

$\displaystyle\sum_{n=1}^N (21n-20) = 2332$

Expand the sum on the left as

$\displaystyle 21 \sum_{n=1}^N n-20\sum_{n=1}^N1 = 2332$

and recall the formulas,

$\displaystyle\sum_{k=1}^n1=\underbrace{1+1+\cdots+1}_{n\text{ times}}=n$

$\displaystyle\sum_{k=1}^nk=1+2+3+\cdots+n=\frac{n(n+1)}2$

So the sum of the first N terms of a (n) is such that

21 × N (N + 1)/2 - 20N = 2332

Solve for N :

21 (N ² + N) - 40N = 4664

21 N ² - 19 N - 4664 = 0

Now for the problem I mentioned at the start: this polynomial has no rational roots, and instead

N = (19 ± √392,137)/42 ≈ -14.45 or 15.36

so there is no positive integer N for which the first N terms of the sum add up to 2332.

4 0

2 years ago

Solve |x| > -9. No solution All reals { x | x < -9 or x > 9}

neonofarm [45]

All reals is the answer.

Since y =-9 graph is way below the absolute value graph. Meaning that y value of modulus is greater than the constant graph.

6 0

3 years ago

Express the following rational number as decimals 615/125

GarryVolchara [31]

Answer:

4.92

Step-by-step explanation:

615 divided by 125 equals to 4.92

4 0

2 years ago

A face of a solid is

nalin [4]

Answer:

In solid geometry, a face is a flat (planar) surface that forms part of the boundary of a solid object; a three-dimensional solid bounded exclusively by faces is a polyhedron. (OR) A face is a 2D shape that makes up one surface of a 3D shape, an edge is where two faces meet and a vertex is the point or corner of a geometric shape.

Step-by-step explanation:

3 0

3 years ago

I know that real numbers consist of the natural or counting numbers, whole numbers, integers, rational numbers and irrational nu

ra1l [238]

The imaginary unit $i$ belongs to the set of complex numbers, denoted by $\mathbb C$ . These numbers take the form $a+bi$ , where $a,b$ are any real numbers.

The set of real numbers, $\mathbb R$ , is a subset of $\mathbb C$ , where each number in $\mathbb R$ can be obtained by taking $b=0$ and letting $a$ be any real number.

But any number in $\mathbb C$ with non-zero imaginary part is not a real number. This includes $i$ .

"is it possible that i can use an imaginary number for a real number"

I'm not sure what you mean by this part of your question. It is possible to represent any real number as a complex number, but not a purely imaginary one. All real numbers are complex, but not all complex numbers are real. For example, 2 is real and complex because $2=2+0i$ .

There are some operations that you can carry out on purely imaginary numbers to get a purely real number. A famous example is raising $i$ to the $i$ -th power. Since $i=e^{i\pi/2}$ , we have

$i^i=\left(e^{i\pi/2}\right)^i=e^{i^2\pi/2}=e^{-\pi/2}\approx0.2079$

3 0

3 years ago