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Could you please Give me examples of arithmetic progressions? - Answers

1,11,21,31,41, ... goes on forever. The common difference is 10. Start with the number 1 and since 10 is the common difference, add 10, you have 11, now add 11 again and you have 21, keep going like this. The sum S of the first n values of a finite sequence is given by the formula: S = 1/2(a1 + an)n, where a1 is the first term and an the last. So in the example I gave, take the first 3 numbers, 1, 11 and 21. S is the sum of them which we know is 33 just by adding 1 +11+21 Now using the formula, S=1/2[(1+21)3]=1/2(66)=33 as we expected. Here is the formula for the nth term an = 1/2(an-1 + an+1)



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Could you please Give me examples of arithmetic progressions? - Answers

https://math.answers.com/other-math/Could_you_please_Give_me_examples_of_arithmetic_progressions

1,11,21,31,41, ... goes on forever. The common difference is 10. Start with the number 1 and since 10 is the common difference, add 10, you have 11, now add 11 again and you have 21, keep going like this. The sum S of the first n values of a finite sequence is given by the formula: S = 1/2(a1 + an)n, where a1 is the first term and an the last. So in the example I gave, take the first 3 numbers, 1, 11 and 21. S is the sum of them which we know is 33 just by adding 1 +11+21 Now using the formula, S=1/2[(1+21)3]=1/2(66)=33 as we expected. Here is the formula for the nth term an = 1/2(an-1 + an+1)



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https://math.answers.com/other-math/Could_you_please_Give_me_examples_of_arithmetic_progressions

Could you please Give me examples of arithmetic progressions? - Answers

1,11,21,31,41, ... goes on forever. The common difference is 10. Start with the number 1 and since 10 is the common difference, add 10, you have 11, now add 11 again and you have 21, keep going like this. The sum S of the first n values of a finite sequence is given by the formula: S = 1/2(a1 + an)n, where a1 is the first term and an the last. So in the example I gave, take the first 3 numbers, 1, 11 and 21. S is the sum of them which we know is 33 just by adding 1 +11+21 Now using the formula, S=1/2[(1+21)3]=1/2(66)=33 as we expected. Here is the formula for the nth term an = 1/2(an-1 + an+1)

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      1,11,21,31,41, ... goes on forever. The common difference is 10. Start with the number 1 and since 10 is the common difference, add 10, you have 11, now add 11 again and you have 21, keep going like this. The sum S of the first n values of a finite sequence is given by the formula: S = 1/2(a1 + an)n, where a1 is the first term and an the last. So in the example I gave, take the first 3 numbers, 1, 11 and 21. S is the sum of them which we know is 33 just by adding 1 +11+21 Now using the formula, S=1/2[(1+21)3]=1/2(66)=33 as we expected. Here is the formula for the nth term an = 1/2(an-1 + an+1)
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