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nydus/The First Steps in AlgebraPublic
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XIII.

A - a = b - A, lintwhence A = a + b2. rintFormula (2)

Hence, the arithmetical mean of any two numbers is found by taking half their sum.

Sometimes it is required to insert several arithmetical means between two numbers.

If m=the number of means, and n=the whole number of terms, then m+2=n. If m+2 is substituted for n in formula (1),

l = a + (n - 1)d, lintthe result is l = a + (m + 1)d. [1] lint By transposing a, l - a = (m + 1) d. l - am + 1 = d. Formula (3)

Thus, if it be required to insert six means between 3 and 17, the value of d is found to be 1736+1=2; and the series will be 3, 5, 7, 9, 11, 13, 15, 17.

Exercise 75.

  1. Find the 25th term in the series 3, 6, 9, .
  1. Find the 13th term in the series 50, 49, 48, .
  1. Find the 15th term in the series 17, 37, 57, .
  1. Find the 19th term in the series 14, 14, 34, .
  1. Find the 10th term in an arithmetical progression whose 1st term is 5 and 3d term 9.
  1. Find the 11th term in an arithmetical progression whose 1st term is 10 and whose 6th term is 5.
  1. If the 3d term of an arithmetical progression is 20 and the 13th term is 100, what is the 20th term?
  1. Which term of the series 5, 7, 9, 11, , is 43?
  1. Which term of the series 43, 32, 53, , is 18?
  1. What is the arithmetical mean of 20 and 32?
  1. What is the arithmetical mean of a+b and ab?
  1. Insert 8 arithmetical means between 20 and 29.

171. To Find the Sum of Any Number of Terms of an Arithmetical Series.

If l denote the last term, a the first term, n the number of terms, d the common difference, and s the sum of the terms, it is evident that the series beginning with the first term will be a, a+d, a+2d, etc., and beginning with the last term will be l, ld, l2d, etc. Therefore,

{r*{12}{c}} s &=& a &+& (a + d) &+& (a + 2d) &+& \dots &+& (l - d) &+& l, \rlap{\quad\text{or}} \ s &=& l &+& (l - d) &+& (l - 2d) &+& \dots &+& (a + d) &+& a \ \hline 2s &=& (a + l) &+& (a + l) &+&

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