How will you get the sum of odd numbers between 240 and 260?

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How will you get the sum of odd numbers between 240 and 260?

What is the sum of odd numbers between 240 and 260? What is the sum of even numbers between 300 and 320?

Research, Knowledge and Information :

How to Add a Sequence of Consecutive Odd Numbers

How to Add a Sequence of Consecutive Odd Numbers. You can add a series of ... if the question asks you to find the sum of all consecutive odd numbers between 1 ...
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Even and Odd Numbers - Math is Fun

Even and Odd Numbers. ... Odd numbers are in between the even numbers. ... When we add (or subtract) odd or even numbers the results are always: Operation
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Formula for consecutive, even, odd integers - Beat The GMAT

... Formula for consecutive, even, odd integers. GMAT Forum. ... let's find the sum of even numbers between 102 to 200. =50 ... (260) ...
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Sum of Consecutive Odd Integers - The Math Forum at NCTM

Sum of Consecutive Odd Integers Date: 07/27/2001 at 05:13:33 From: Hooji F. Rubie Subject: Number as a sum of consecutive odd integers? Given an integer N, I would ...
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What is the sum of the first 100 odd numbers? |

What is the sum of the first 100 odd numbers? A: Quick Answer. The sum of the first 100 odd numbers is 10,000.
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Parity (mathematics) - Wikipedia

Thus, there is here between odd and even numbers one number (one) which is neither of the two. ... usually defined as the parity of the sum of the coordinates.
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Learning Math: Even and Odd Numbers - Syvum Technologies

The odd numbers between 30 and 40 are 31, 33, ... Their sum is (80 × 3) + (2 + 4 + 6) = 240 + 12 = 252. 22. What is the sum of the ... Learning Math: Even and Odd ...
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What is the sum of 2+4+6+8+…+98+100? - Quora

Here is a general equation for the sum of consecutive numbers. [math]\frac{n(a_1 + a_n)}{2}[/math] [math]a[/math] represents the member of the set of numbers you're ...
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Suggested Questions And Answer :

how many ways are there to add and get the sum of 180

There are an infinite number of ways to get 180 from two numbers, if we count decimals and fractions as well as other real numbers; but if we are limited to positive integers greater than zero and just the sum of two of them, we are limited to x and 180-x. If we also exclude 90+90 because
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If 1=3,2=5,3=6,4=9 so how much word needed for equal of 5?

5=10 using the following logic. Switch to the binary system of counting: 1, 10, 11, 100, 101, 110, 111, 1000, 1001, 1010 represent the numbers 1 to 10 decimal in binary. If we cross out those numbers with an odd number of ones we get: 11, 101, 110, 1001, 1010. When these are converted bac
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Find the explicit expression for the sequence: 1.5, 1.6, 1.8, 2.1, 2.5

The common difference of this series is 0.1n, because the differences are 0.1, 0.2, 0.3, 0.4. a1=a0+0.1, a2=a1+0.2, a3=a2+0.3,..., a(n)=a(n-1)+0.1n. If we subtract 1.5 from each term, we get: 0, 0.1, 0.3, 0.6, 1.0. This is 0, 0+0.1, 0+0.1+0.2, 0+0.1+0.2+0.3, 0+0.1+0.2+0.3+0.4. This is the same
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10 vedic maths rules for class IX

2 instead of 5: 34/5 can be calculated by multiplying 34 by 2 instead of dividing by 5. 34*2=68. We just move the decimal point one place to the left: 34/5=6.8. 124/5=24.8 because 124*2=248. Move the decimal point: 248 becomes 24.8. 34*5 is the same as 34/2=17 but we add a zero to make 17 into
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Need help with word problem

Let the age of one surfer be 10a+b, then the age of the other is 10b+a. Add them together: 11a+11b, so 1/11 is a+b=sqrt(10b+a-10a-b)+1=sqrt(9b-9a)+1=sqrt(9(b-a))+1=3sqrt(b-a)+1. If b-a is to be a perfect square, the only ones are 0, 1, 4, 9, making a+b=1, 4, 7, 10. Now there's a problem: if we
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find HCF and LCM of3x^4+17x^3+27x^2+7x-6;6x^4+7x^3-27x^2+17x-3 by division method.

3x^4+17x^3+27x^2+7x-6=(x+1)(3x^3+14x^2+13x-6)=(x+1)(x+3)(3x^2+5x-2)= (x+1)(x+3)(3x-1)(x+2) CLUE: Write coefficients: 3 17 27 7 -6. If we add them together, do we get zero? No, so 1 is not a zero and x-1 is not a factor. Look at the odd powers of x (17x^3 and 7x). Change the sign in fro
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find all reconstuctions of the sum: ABC+DEF+GHI=2014 if all letters are single digits

There is no solution if A to I each uniquely represent a digit 1 to 9, because the 9's remainder of ABC+DEF+GHI (0) cannot equal the 9's remainder of 2014 (=7). Let me explain. The 9's remainder, or digital root (DR), is obtained by adding the digits of a number, adding the digits of the result, an
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How many different distributions can the manager make if every employee receives at least one voucher?

There must be 100 vouchers because each is worth RM5 and the total value is 500 ringgits or RM500. (i) Each employee receives at least 1 voucher, so that means there are 95 vouchers left to distribute. We can write each distribution as {A,B,C,D,E}: starting with {0,0,0,0,95}, then {0,0,0,1,94}
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0,2,4,4,6,8,9,11,13,15 what are the five numbers?

Call the 5 numbers A, B, C, D and E. We assume these to be integers, since there's no indication otherwise. The sums are A+B, A+C, A+D, A+E, B+C, B+D, B+E, C+D, C+E, D+E. Let's look at the sum zero. To get the sum of two number to be zero, they're either both zero or one is the negative of the
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for any n belongs N define Pn as the set containing all primes that are the sum of n consecutive primes

The first few primes are 1, 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47 (P<50). The sum of consecutive primes that are themselves are primes can be identified: p1=1; p2=1+2=3; p4=1+2+3+5=11; p6=1+2+3+5+7+11=29; p8=1+2+3+5+7+11+13+17=59; p10=1+2+3+...+17+19+23+101, ... My under
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