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Example 1: Find the 10th term and the nth term for the sequence 7, 10, 13, … . Solution: Un= U10 =

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Example 2 Find the three numbers in an arithmetic progression whose sum is 24 and whose product is 480. Solution Let the three numbers be (a-d), a, (a+d) where d is the common difference. (a-d)(a)(a+d)=480 (a-d) +a + (a+d) = 24 Substituting a =8, we get 3a = 24 a = 8

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**When d = 2, required numbers are 6, 8, 10**

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Example 3 The sum of the first eighteen terms of an arithmetic series is -45 and the eighteenth term is also -45. Find the common difference and the sum of the first hundred terms. Solution (1)

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(2) Solving equation 1 and 2, we get

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Example 4 Find the sum of the positive integers which are less than 500 and are not multiples of 11. Solution Sum of multiples of 11(SII) Sum of integers from 1 to 499 (SI) Required sum=

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Required sum = SI-SII = =113365

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Example 5 The sum of the first 10 terms of an A.P. is 145 and the sum of the next 6 terms is 231 Find (i) the 31st term , and (ii) the least number of terms required for the sum to exceed 2000.

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Solution (1) (2)

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(1)-(2): From (2): (ii)Find the least number of terms required for the sum to exceed 2000. Let n be the least number of terms required for

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Consider

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For + - + Since n must be a positive integer, n > 36.7 Hence, the least number of terms required is 37.

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Example 6 Given that the fifth term of a geometric progression is and the third term is 9. Find the first term and the common ratio if all the terms in the G.P. are positive. (1) (2)

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(1)(2): Since all the terms in the G.P. are positive, r > 0 From (2):

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Example 7 Three consecutive terms of a geometric progression are and 81. Find the value of x. If 81 is the fifth term of the geometric progression, find the seventh term.

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**Given 81 is the fifth term, find the seventh term:**

U5= ar4 = 81 U7 = ar6

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Example 8 Find the sum of the first eight terms of the series Soln: G.P. with

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Example 9 A geometric series has first term 1 and the common ratio r, where r 1, is positive. The sum of the first five terms is twice the sum of the terms from the 6th to 15th inclusive. Prove that

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Solution Given Since and a = 1, we have

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Let So the equation becomes Let f(x) = Since f(1) = 2 – = 0 (x –1) is a factor of f(x). Comparing coefficient of x:

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Hence, (r<0) Since and r > 0,

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Sum to infinity, S (or ) = The sum to infinity exists (the series converges or series is convergent) provided Example 10 Determine whether the series given below converge. If they do, give their sum to infinity. G.P. with >1 (a) Does not converge

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G.P. with (b) <1 Series converges

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G.P. with (c) >1 Does not converge

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Example 11 A geometric series has first term a and the common ratio Show that the sum to infinity of the geometric progression is Solution (shown)

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Example 12 A geometric series has first term a and common ratio r. S is the sum to infinity of the series, T is the sum to infinity of the even-numbered terms (i.e ) of the series. Given that S is four times the value of T, find the value of r. common ratio r 2 Given S = 4T, we have

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©Brooks/Cole, 2001 Chapter 12 Derived Types-- Enumerated, Structure and Union.

©Brooks/Cole, 2001 Chapter 12 Derived Types-- Enumerated, Structure and Union.

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