# Geometric Sequences and Sums

 Tutorial on geometric sequences and summations. A - Geometric Sequences An arithmetic sequence is a sequence of numbers that is obtained by multiplying the preceding number by a constant number called the common ratio. Example 1: 1,2, 4, 8, 16, ... each term of the sequence is obtained by multiplying by 2 the preceding term. If a1 is the first term and r is the common difference, the nth term an is given by an = a1 × rn-1 The sum Sn of the first n terms of an arithmetic sequence is given by Sn = a1(1 - rn) / (1 - r) If the common ration r is such that |r| < 1 and the sum is infinite, then the sum is given by S∞ = a1 / (1 - r) Examples with Detailed Solutions Example 3: The 4th and 7th terms of a geometric sequence are 1/8 and 1/64 respectively. Find the 9th term. Solution to Example 2: a4 = a1 r 4-1 = a1 r 3 = 1/8 a7 = a1 r 7-1 = a1 r 6 = 1/64 Then: a7 / a4 = a1 r 6 / a1 r 3 = r 3 = (1/64) / (1/8) = 1/8 Hence r = 1/2 The 9th term: a9 = a1 r 9-1 = a1 r 8 = a1 r 6 × r 2 = a7 (1/2) 2 = (1/64)(1/4) = 1/256 Example 3: Find 3 consecutive numbers in a geometric sequence whose sum is 234 and their product is 157,646. Solution to Example 3: Let the 3 terms be : x , x r and x r 2 sum = x + x r + x r 2 = x(1 + r + r 2) = 234 product = x(x r)(x r 2) = x 3 r 3 = 157464 The last equation gives: x r = 3 √ (157464) = 54 or x = 54/r We now substitute x by 54/r in the equation x(1 + r + r 2) = 234 obtained above to get (54/r)(1 + r + r 2) = 234 or 54(1 + r + r 2) = 234 r Solve the above quadratic equation for r to obtain r = 3 or r = 1/3 Two possible solutions to the given problem: Use r = 3 and x r = 54 to find the three terms: x = 18, x r = 54 and x r 2 = 162 Use r = 1/3 and x r = 54 to find the three terms: x = 162, x r = 54 and x r 2 = 18 Example 4: The first term of a geometric sequence is 2000 and the second term is 1000. Starting from the first term, how many consecutive terms in this sequence must be taken in order to obtain a sum equal to 3875? Solution to Example 4: First the common ration r = 1000/2000 = 1/2 In a geometric sequence the sum is given by Sn = a1(1 - rn) / (1 - r) = 2000(1 - (1/2)n) / (1 - 1/2) = 3875 We rewrite the above equation as follows: 1 - (1/2)n = ( 3875 × (1/2) )/ 2000 which gives (1/2)n = 1 - ( 3875 × (1/2) )/ 2000 = 125 / 4000 = 1/32 = (1/2)5 Which gives n = 5. The sum of the first five terms gives 3875 Example 5: Prove that x, x2 + 1 and x3 + x cannot be the 3 consecutive terms in a geometric sequence of real numbers. Solution to Example 5: Suppose they are the three terms are that of a geometric sequence and express the common ratio using the three terms and write the following equation (x2 + 1) / x = ( x3 + x ) / ( x2 + 1 ) The cross product gives (x2 + 1) (x2 + 1) = x ( x3 + x ) Expand and simplify 2 x2 + 1 = 0 The above equation has no real solution and therefore the sequence as defined above cannot be a geometric sequence with real numbers. Example 6: The first three terms of an arithmetic sequence are as follows: x , x2 + 4 and 16x. Find the three terms. Solution to Example 6: Use the three terms to find two expressions of the common ratio and equate them to obtain the equation (x2 + 4) / x = 16 x / x2 + 4 Cross product gives (x2 + 4)2 = 16 x2 Expand and simplify x4 + 8x4 + 16 = 16 x2 x4 - 8x4 + 16 = 0 (x2 - 4)2 = 0 x2 - 4 = 0 x = 2 or x = -2 Two possible solutions: 1) x = 2, Hence the three terms are :x = 2, x2 + 4 = 8 and 16 x = 32 2) x = -2, Hence the three terms are :x = -2, x2 + 4 = 8 and 16 x = - 32
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Updated: 21 October 2014 (A Dendane)