Subject: sequence and series
Name: Michael

Who is asking: Student

Level: All

Question:

the sum of the first ten terms of an arithmetic series is 100 and the first term is 1. Find the 10th term.

the common ratio in a certain geometric sequence is r=0.2 and the sum of the first four terms is 1248 find the first term.

Hi Michael

There are formulas to do these problems but I don't remember them and I encourage my students not to remember them.

What I remember about an arithmetic series is that if you have to sum it do it once forward and once backward. So in your case where the sum of ten terms is 100,

a | + | (a + d) | + | ... | + | (a + 8d) | + | (a + 9d) | = | 100 | and |

(a + 9d) | + | (a + 8d) | + | ... | + | (a + d) | + | a | = | 100 | |

Adding these two equations together gives:
(2a + 9d) | + | (2a + 9d) | + | ... | + | (2a + 9d) | + | (2a + 9d) | = | 200 |

That is 10(2a + 9d) = 200 or 2a + 9d = 20. Since you know that a = 1 this gives 2 + 9d = 20 and hence d = 2.
What I remember about the sum of a geometric series is to multiply both sides by the common ratio r. For your problem,

a | + | ar | + | ar^{2} | + | ar^{3} | | | = | 1248. | Multiplying both sides by r gives |

| | ar | + | ar^{2} | + | ar^{3} | + | ar^{4} | = | 1248r | |

(After multiplying the left side by r I shifted it so that like terms were lined up.) In this case subtract the two equations to get,
a - ar^{4} = 1248 - 1248r. Substituting r = 0.02 gives a = 1000.

Cheers,

Penny