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try findin the answer in r.s aggarwal

Posted on Mar 11, 2008

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Posted on Jan 02, 2017

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Are you trying to find a particular term in the geometric sequence?

For example, in the geometric sequence 2,4,8,16 to find the 15th term we would use the formula a sub n = a sub 1 x r ^(n-1), where a sub n is the nth term of the geometric sequence

a sub 1 is the 1st term of the geometric sequence

r is common ratio between successive terms

n is the term you are looking for

In our example, a sub 1 is 2, r is 2 and n is 15.

a sub n = a sub 1 x r ^(n-1)

a sub 15 = 2x 2(15-1)

a sub 15 = 2 ^ 15

a sub 15 = 32,768

On your calculator, you use the key to the right of x^2 to do x to the nth power.

Good luck.

Paul

Geometric Sequences and Series

casio fx 300es manual Google Search

For example, in the geometric sequence 2,4,8,16 to find the 15th term we would use the formula a sub n = a sub 1 x r ^(n-1), where a sub n is the nth term of the geometric sequence

a sub 1 is the 1st term of the geometric sequence

r is common ratio between successive terms

n is the term you are looking for

In our example, a sub 1 is 2, r is 2 and n is 15.

a sub n = a sub 1 x r ^(n-1)

a sub 15 = 2x 2(15-1)

a sub 15 = 2 ^ 15

a sub 15 = 32,768

On your calculator, you use the key to the right of x^2 to do x to the nth power.

Good luck.

Paul

Geometric Sequences and Series

casio fx 300es manual Google Search

Jan 12, 2016 | Casio fx-300ES Calculator

Let m represent the subscription rate for a man, and w the subscription rate for a woman.

m/w=4/3, or** m=(4/3)w**.

2 men and 5 women paid 460. Thus

**2m+5w=460.**

Substitute**(4/3)w** for **m**, in **2m+5w=460**.

This gives an equation in the unknown w. Solve for w

2(4/3)w +5w=460 or (23/3)w=460.**Answer is w=60**

Use this value in the relation m=(4/3)w to get m=80.

**Check m/w=80/60=4/3**

m/w=4/3, or

2 men and 5 women paid 460. Thus

Substitute

This gives an equation in the unknown w. Solve for w

2(4/3)w +5w=460 or (23/3)w=460.

Use this value in the relation m=(4/3)w to get m=80.

May 09, 2014 | The Learning Company Achieve! Math &...

Assuming -31 to be the first term and not the zeroth term, the nth term is 3n-34. This makes 92 the 42nd term.

Mar 07, 2013 | Cars & Trucks

Here, We deal with Some Special Products in Polynomials.

Certain products of Polynomials occur more often

in Algebra. They are to be considered specially.

These are to be remembered as Formulas in Algebra.

Remembering these formulas in Algebra is as important

as remembering multiplication tables in Arithmetic.

We give a list of these Formulas and Apply

them to solve a Number of problems.

We give Links to other Formulas in Algebra.

Here is the list of Formulas in

Polynomials which are very useful in Algebra.

Formulas in Polynomials :

** Algebra Formula 1 in Polynomials: ** * Square of Sum of Two Terms: *

** (a + b)2 = a2 + 2ab + b2 **

** (a - b)2 = a2 - 2ab + b2 **

** (a + b)(a - b) = a2 - b2 **

** (a + b)(a2 - ab + b2) = a3 + b3 **

** (a - b)3 = a3 - 3a2b + 3ab2 - b3 = a3 - 3ab(a - b) - b3 **

(First term + Second term)2

= (First term)2 + 2(First term)(Second term) + (Second term)2

Jul 02, 2011 | Computers & Internet

This sum is equal to 40+48+56+64+72+...+3024+3032+3040+3048= 40+(48+3048)+(56+3040)+(64+3032)+(72+3024)...

There are (3048-40)/8+1=377 terms in sum. (48+3048)=(56+3040)=(64+3032)=(72+3024)=3096 and there are (377-1)/2=188 pairs whose sum is 3096. Only term that is not in such pair is 40.

Finally sum is 40+188*3096=582088.

There are (3048-40)/8+1=377 terms in sum. (48+3048)=(56+3040)=(64+3032)=(72+3024)=3096 and there are (377-1)/2=188 pairs whose sum is 3096. Only term that is not in such pair is 40.

Finally sum is 40+188*3096=582088.

Jul 01, 2011 | Computers & Internet

Using elementary algebria in the **binomial theorem, **I expanded the power **(***x* + *y*)^n into a sum involving terms in the form a x^b y^c. The coefficient of each term is a positive integer, and the sum of the exponents of *x* and *y* in each term is **n**. This is known as binomial coefficients and are none other than combinatorial numbers.

**Combinatorial interpretation:**

Using** binomial coefficient (n over k)** allowed me to choose** ***k* elements from an **n**-element set. This you will see in my calculations on my Ti 89. This also allowed me to use **(x+y)^n** to rewrite as a product. Then I was able to combine like terms to solve for the solution as shown below.

(x+y)^6= (x+y)(x+y)(x+y)(x+y)(x+y)(x+y) = x^6+6x^5y+15x^4y^2+20x^3y^3+15x^2y^4+6xy^5+y^6

**This also follows Newton's generalized binomial theorem:**

**Now to solve using the Ti 89.**

Using

(x+y)^6= (x+y)(x+y)(x+y)(x+y)(x+y)(x+y) = x^6+6x^5y+15x^4y^2+20x^3y^3+15x^2y^4+6xy^5+y^6

**Using sigma notation, and factorials for the combinatorial numbers, here is the binomial theorem:**

**The summation sign is the general term. Each term in the sum will look like that as you will see on my calculator display. Tthe first term having k = 0; then k = 1, k = 2, and so on, up to k = n. **

Jan 04, 2011 | Texas Instruments TI-89 Calculator

You did not provide a general (n) term. However, I figured it to be something involving n^2. Hence the sum to n terms can be of the form: xn^3 + yn^2 + zn. Using the partial sums: 1, 6 (1+5), and 18 (1+5+12) we can build a linear equations system:

1= x + y +z (n=1)

6 = 8x + 4y +2z (n=2, n^2=4, n^3 = 8)

18 = 27x + 9y + 3z (n=3, n^=9, n^=27)

Solving for x,y,z we get x = 1/2, y= 1/2, z =0

Hence 1+5+12+22+..+ f(n^2)= 1/2(n^3+n^2)

1= x + y +z (n=1)

6 = 8x + 4y +2z (n=2, n^2=4, n^3 = 8)

18 = 27x + 9y + 3z (n=3, n^=9, n^=27)

Solving for x,y,z we get x = 1/2, y= 1/2, z =0

Hence 1+5+12+22+..+ f(n^2)= 1/2(n^3+n^2)

Nov 04, 2010 | Refrigerators

Hello,

I start with the most complicated

Right hand side

2 ( 9 +6m-3+m)=2(7m +6) =14m +12

RHS= 14m +12

Now it is your turn to rearrange the left hand side, cast it into a sum of at most 2 terms, one containing m the other not.

LHS= ?

Once you find that you rewrite LHS=RHS, then put all the terms with m on the left, changing sign if a term changes side; put all the terms witout m on the right, changing sign if the term changes side.

Once you finish with that, if the sole term with m on the left is multiplied by a factor other than 1, you must divide both sides of the equality by that factor to isolate m.

Now it is up to you. I am more interested in helping you find the solution that chew it for you. Learning is like eating:Nobody can do that for you.

Hope it helps.

I start with the most complicated

Right hand side

2 ( 9 +6m-3+m)=2(7m +6) =14m +12

RHS= 14m +12

Now it is your turn to rearrange the left hand side, cast it into a sum of at most 2 terms, one containing m the other not.

LHS= ?

Once you find that you rewrite LHS=RHS, then put all the terms with m on the left, changing sign if a term changes side; put all the terms witout m on the right, changing sign if the term changes side.

Once you finish with that, if the sole term with m on the left is multiplied by a factor other than 1, you must divide both sides of the equality by that factor to isolate m.

Now it is up to you. I am more interested in helping you find the solution that chew it for you. Learning is like eating:Nobody can do that for you.

Hope it helps.

Oct 09, 2009 | Texas Instruments Office Equipment &...

Nth Term = _______________

3

(-1 + n)

put n = 0 in the above equation, you will get 10. Put n = 1 and the result is 22, and so on.

Hope this answers your question!

Oct 05, 2009 | The Learning Company Treasure MathStorm!

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