By Abraham P Hillman

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**Additional info for Algebra through problem solving **

**Sample text**

G3m be a geometric progression of positive terms. Let A, B, and C be the geometric means of the first m terms, the first 2m terms, and all 3m terms, respectively. Show that B2 = AC. 26 35. , 30 which are divisible exactly by 3 or 5 (or both), and let T consist of those divisible by neither 3 nor 5. (a) Write out the sequence of numbers in S in their natural order. (b) In the sequence of Part (a), what is the arithmetic mean of terms equally spaced from the two ends of the sequence? (c) What is the arithmetic mean of all the numbers in T?

In addition, the former side AB has become a diagonal of the new (k + 1)-sided polygon. Thus a (k + 1)-sided polygon has a total of k(k&3) % (k & 2) % 1 diagonals. But: 2 k(k&3) % (k & 2) % 1 2 k 2 &3k % 2k & 2 ' 2 30 k2 & k & 2 2 (k % 1)(k & 2) ' 2 (k % 1)[(k % 1) &3] ' 2 ' This is the desired formula for n = k + 1. So, by assuming that the formula Dn = n(n - 3)/2 is true for n = k, we have been able to show it true for n = k + 1. This, in addition to the fact that it is true for n = 3, proves that it is true for all integers greater than or equal to 3.

60,000 of all the positive integers not exceeding 60,000 which are integral multiples of at least one of 4, 5, and 6. 37. , t - 2. Show that the t terms are in arithmetic progression. 38. Find a compact expression for the sum v1 + v2 + ... + vt in terms of v1 and v2, given that vn+2 = (vn+1)2/vn for n = 1, 2, ... ,t - 2. 39. Let an = 2n be the nth term of the geometric progression 2, 22, 23, ... , 2t. Show that an+2 - 5an+1 + 6an = 0 for n = 1, 2, ... , t - 2. 40. For what values of r does the sequence bn = rn satisfy bn+2 - 5bn+1 + 6bn = 0 for all n?

### Algebra through problem solving by Abraham P Hillman

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