Finding the Smallest Positive Term in an Arithmetic Series (C1 Level)

  • Thread starter Thread starter CathyLou
  • Start date Start date
  • Tags Tags
    Arithmetic Series
AI Thread Summary
To find the smallest positive term in an arithmetic series, one can rearrange the formula S_n = n[2a_1 + (n-1)d]/2, where S_n represents the sum of the first n terms, n is the number of terms, a_1 is the first term, and d is the common difference. This formula allows for the calculation of terms in the series based on the initial term and the common difference. Understanding how to manipulate this equation is crucial for determining specific values within the series. The discussion emphasizes the importance of recognizing the components of the formula for effective problem-solving. Mastery of this formula is essential for success at the C1 level in mathematics.
CathyLou
Messages
173
Reaction score
1
Please could someone tell me the way to find the smallest positive term of an arithmetic series (C1 level) as I cannot find a formula anywhere.

Thank you.

Cathy
 
Physics news on Phys.org
You could rearrange the formula S_n=\frac{n[2a_1+(n-1)d]}{2}, where n is the numbers of terms, Sn is the sum of the first n terms, d is the difference between the ith and (i+1)th term, and a1 is the first term of the series.
 
cristo said:
You could rearrange the formula S_n=\frac{n[2a_1+(n-1)d]}{2}, where n is the numbers of terms, Sn is the sum of the first n terms, d is the difference between the ith and (i+1)th term, and a1 is the first term of the series.

Okay. Thanks for your help.

Cathy
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
Back
Top