# Maths the ratio of the sums of m and n terms of an A.P is M^2:N^2.show that the ratio of the Mth and Nth term is (2M-1):(2N-1).

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try findin the answer in r.s aggarwal

Posted on Mar 11, 2008

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## Related Questions:

### How to put geometric sequence in this calculator?

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

Jan 12, 2016 | Casio fx-300ES Calculator

### The subscription fee to a club for men and women are in the ratio 4:3. Two men and five women pay a total sum of \$460. What is the subscription fee for each man?

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

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

### Wat is special product

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
Algebra Formula 2 in Polynomials:

Square of Difference of Two Terms:

(a - b)2 = a2 - 2ab + b2
Algebra Formula 3 in Polynomials:

Product of Sum and Difference of Two Terms:

(a + b)(a - b) = a2 - b2
Algebra Formula 4 in Polynomials:

Product giving Sum of Two Cubes:

(a + b)(a2 - ab + b2) = a3 + b3
Algebra Formula 5 in Polynomials:

Cube of Difference of Two Terms:

(a - b)3 = a3 - 3a2b + 3ab2 - b3 = a3 - 3ab(a - b) - b3
Algebra Formula 8 in Polynomials:

Each of the letters in fact represent a TERM.

e.g. The above Formula 1 can be stated as
(First term + Second term)2
= (First term)2 + 2(First term)(Second term) + (Second term)2

Jul 02, 2011 | Computers & Internet

### Find the sum of all integer multiples of 8 between 33 and 3051.

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.

Jul 01, 2011 | Computers & Internet

### TI-89 Titanium: I want to solve a Binomial Theorem problem (x+y)^6 how would i go about solving this in the calculator?

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 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.
Notice that the sum of the exponents (n ? k) + k, always equals n.

The summation being preformed on the Ti 89. The actual summation was preformed earlier. I just wanted to show the symbolic value of (n) in both calculations. All I need to do is drop the summation sign to the actual calculation and, fill in the term value (k), for each binomial coefficient.

This is the zero th term. x^6, when k=0. Notice how easy the calculations will be. All I'm doing is adding 1 to the value of k.

This is the first term or, first coefficient 6*x^5*y, when k=1.
Solution so far = x^6+6*x^5*y

This is the 2nd term or, 2nd coefficient 15*x^4*y^2, when k=2.
Solution so far = x^6+6*x^5*y+15*x^4*y^2

This is the 3rd term or, 3rd coefficient 20*x^3*y^3, when k=3.
Solution so far = x^6+6*x^5*y+15*x^4*y^2+20*x^3*y^3

This is the 4th term or, 4th coefficient 15*x^2*y^4, when k=4.
Solution so far = x^6+6*x^5*y+15*x^4*y^2+20*x^3*y^3+15*x^2*y^4

This is the 5th term or, 5th coefficient 6*x*y^5, when k=5.
Solution so far = x^6+6*x^5*y+15*x^4*y^2+20*x^3*y^3+15*x^2*y^4+6*x*y^5

This is the 6th term or, 6th coefficient y^6, when k=6.
Solution so far = x^6+6*x^5*y+15*x^4*y^2+20*x^3*y^3+15*x^2*y^4+6*x*y^5+y^6

Putting the coefficients together was equal or, the same as for when I used the expand command on the Ti 89.

binomial coefficient (n over k) for (x+y)^6
x^6+6*x^5*y+15*x^4*y^2+20*x^3*y^3+15*x^2*y^4+6*x*y^5+y^6

Jan 04, 2011 | Texas Instruments TI-89 Calculator

### Sum to n terms of the series 1+5+12+22..........

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)

Nov 04, 2010 | Refrigerators

### 5(2m+3)-(1-2m)=2[3(3+2m)-(3-m)] What is m in this problem?

Hello,
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 question

(-10 - 4 (-2 + n) n)
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.

Oct 05, 2009 | The Learning Company Treasure MathStorm!

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