CPT Section D - Quantitative Aptitude Chapter 6 CA. Loveneesh Kapoor
Learning Objectives Understand the concept of Sequence and Series Understand the nature of sequences-arithmetic Progression (A.P.) and Geometric Progression (G.P.) To find any term of the sequence or the sum of all terms in the sequence Understand various formulas of A.P and G.P series. Numerical as well as mathematical problems related to the A.P and G.P Series.
Sequences-Pattern Guess the next number in the following cases 1,4,7,10,, Next no. will be 13 and then 16 following a definite order of succession of 3. 5,15,25,, Next term will be 35, adding definite succession of 10 to the previous number 25. Now Next Number can be judged because there is a definite pattern and the sequence is worked out with a common difference among all the numbers. This succession is called as ARITHMETIC PROGRESSION (A.P.)
Sequence-Meaning A succession of terms arranged in a definite order and following some definite rule or law. In other words, numbers or objects arranged in a definite order. Example 1,3,5,7,9,.. Odd Numbers 2,4,6,8,10..Even Numbers
Arithmetic Progressions (A.P. Series)
nth term of the series 5, 10, 15, 20, 25,..next term will be 30 following pattern of succession of 5. t 1 = 5 = 5 + 0(5) = a + 0 (d) t 2 = 10 = 5 + 1(5) t 3 = 15 = 5 + 2(5) = a + 1 (d) = a + 2 (d) t n = 5 + (n-1)(5) = a + (n-1)d Thus, n th term i.e, t n = a + (n-1)d & the series is a, a+d, a+2d, a+3d..a+(n-1)d
Example-1 Q : 1,4,7,10, Find the 20 th term. Sol: Here a = 1, d = 4-1 = 3, n=20 t n = a + (n-1) d t 20 = 1 + (20-1) 3 = 1 + 19 (3) = 1+57 = 58
Example-2(a) Q : 12,24,36,.. Find t 14. Solution: a=12, d= 24-12=12, n =14 t n = a + (n-1)d t 14 = 12+(14-1)12 = 12+13(12) = 12+156 =168
Example-2 (b) Q : 12,24,36,.. Which term is 228. Sol: a=12, d= 24 12 = 12, t n = 228 =t n = a + (n-1)d =t n = 12+(n-1)12 = 12+(n-1)12 =228 = (n-1)12 = 228-12 = (n-1)12 = 216 = (n-1) = 216/12 = n-1 = 18 = n = 18+1 = n = 19 Thus 228 is 19 th term of the series.
Example-3 Q: If 5th and 12th terms of an A.P. are 14 and 35 respectively, find the A.P. Sol: Here t 5 = 14, t 12 = 35 t 5 = a + 4d =14 t 12 = a + 11d =35 Solving we get d = 3 and a = 2 Thus the series 2, 5, 8,.
Example 4 Q : How many terms are there in the series -3,3,9, 117 Sol: Here a= -3, d = 3 - (-3) = 6 and last term =117 Last term =t n = a+(n-1)d = 117 = 117 = -3+ (n-1)6 =117+3 =(n-1)6 = 120 =(n-1)6 =120/6 =(n-1) =20+1 = n =21 =n Thus 117 is t 21 or 21 st term
Arithmetic Mean In case of AP series all the numbers lying between 1st and the last term are Arithmetic Means. i.e, A.M= (a + b) / 2 In case of AP series all the numbers lying between 1 st and the last term are Arithmetic Means. Example : The Arithmetic Mean between 8 & 48 is calculated as (8 + 48)/ 2 = 28. Thus 8, 28, 48 is an A.P.
Example 5 Q : Insert 4 AMs between 4 & 324. Sol: A.P. Series 4, --, --, --, --, 324 Here t 1 = a = 4 t 6 = a + 5d = 324 Now, t 6 = 4 + 5d = 324 = 5d=324-4 = 5d=320 = d =320/4 = d =64 Thus, t 2 = a + d = 4 + 64 = 68 t 3 = a + 2d= 4 +2(64) = 132 t 4 = a +3d = 4 + 3(64)=196 t 5 = a+4d = 4 + 4(64) = 260
Q : Insert 4 AMs between 4 & 324. Sol: A.P. Series 4, --, --, --, --, 324 Here t 1 = a = 4 t 6 = a + 5d = 324 Now, t 6 = 4 + 5d = 324 = 5d=324-4
Example 6 Q : Find three terms of AP such that their sum is 33 and product is 1155. Sol: Let the three terms are (a-d), (a), (a+d) Sum = a - d + a + a + d=33 3a = 33 a =11 Now, (a-d).(a).(a+d) =1155 (11-d).11.(11+d) = 1155 (11 2 d 2 ) =1155/11 121-d 2 =105 16 = d 2 ± 4 = d Thus if d = 4, numbers are 7,11, 15 & if d =- 4, numbers are 15, 11, 7
Example 7 Q : Find the number of terms between 74 & 25556 divisible by 5. Sol : Between 74 & 25556 the series of numbers divisible by 5 are forming an AP series 75, 80, 85,..25555 Now 25555 is the nth term of the series t n = a + (n-1)d = 25555 75+ (n-1)5 = 25555 (n-1)5 = 25555-75 (n-1)5 = 25480 (n-1) = 25480/5 n = 5096+1 = 5097 Thus there are 5097 terms of in the AP series.
Example 8(a) Q : Find the 1 st three terms of the series t n = n 2 2n Sol : The series can be formed by putting 1,2,3 in place of n in the given function t 1 = (1) 2 2(1) = 1-2 = -1 t 2 = (2) 2 2(2) = 4-4 = 0 t 3 = (3) 2 2(3) = 9-6 = 3 Thus the three terms are -1, 0, 3
Example 8 (b) Q : Find the nth term of the series 1,3,5,7,. Sol : Here a =1, d = 3-1 = 2 nth term = t n = a + (n-1)d = 1 + (n-1)2 = 1 + 2n-2 = 2n-1 Thus nth term is expressed by the function 2n-1
Sum of A.P. Series In an Arithmetic Progression (A.P.) Series, the sum of all the terms upto n terms has been formulated as S n = n/2{2a+ (n-1)d} Here a is the 1 st term d is the common difference n is the no. of terms Also S n = n/2 [a + l] a is the 1 st term of the series l is the last term of the series
Example 9 Q : Find the sum upto 50 terms of the series2,4,6,8,. Sol Here a = 2 d = 4-2 =2, n = 50 S n = n/2{2a+ (n-1)d} = 50/2[ 2*2 + (50-1)2] =25[4+49*2] = 25[4+98] = 25[102] = 2550
Example 10 Q : Find S 100 for the series 9, 5, 1,. Sol : Here a = 9, d = 5-9 = -4, n = 100 S n = n/2{2a+ (n-1)d} S 100 = 100/2[ 2*9 +(100-1) -4] = 50[18+99*(-4)] = 50[18-396] =50[-378] =(18900)
Example 11 Q : The 1 st and the last term of an A.P. series are -4 and 146 respectively. The sum of terms is 7171, find the number of terms. Sol : Here a= -4, l = 146, S n = 7171 S n = n/2[a + l] = 7171 = n/2 [-4+146] =7171 = n/2[142] =7171 = 71n =7171 =n =7171/71= 101 Thus the number of terms are 101
MCQ-1 Q : The Sum of all natural numbers between 100 and 1000 which are multiples of 5 is a)98450 b) 96450 c) 97450 d) 95450 Answer : (a) 98450 Hint : a=105, d = 5, t n = 995
MCQ-2 Q : Insert 4 AMs between 3 and 18 a) 12,15,9,6 b) 6,9,12,15 c) 9,6,12,15 d) 15,12,9,6 Answer : (b) 6,9,12,15 Hint : a=3, t 6 = 18
MCQ-3 Q : Find the sum of series 8+3-2-7..20 terms a)-790 b) 790 c) -970 d) 970 Answer : (a) -790 Hint : a=8, d= -5, n = 20
MCQ-4 Q : The value of x such that 8x +4, 6x-2, 2x+7 will form an A.P a) 15 b) 2 c) 15/2 d) None of these Answer : (c) 15/2 Hint : d = t 2 t 1 = t 3 t 2
MCQ-5 Q : The sum of three integers in AP is 15 and their product is 80. The integers are a) 2,8,5 b) 8,2,5 c) 2,5,8 d)5,8,2 Answer : (c) 2,5,8 Hint : Let the three terms are (a - d), a, (a + d)
MCQ-6 Q : The sum of a certain number of terms of an AP series -8,-6,-4,..52. The number of terms is a) 12 b) 13 c) 11 d) 14 Answer : (b) 13 Hint : a= -8, d = +2, S n =52
MCQ-7 Q : The sum of a 1 st n odd numbers is a) n 2 b) (n-1) 2 c) (n+1) 2 d) n Answer : (a) n 2 Hint : a= 1, d = 2
MCQ-8 Q : The last term of the AP is 0.6, 1.2, 1.8,. to 13 terms is a) 8.7 b) 7.8 c) 7.7 d) 8.8 Answer : (b) 7.8 Hint : a= 0.6, d = 0.6
MCQ-9 Q : The sum of all natural numbers between 500 and 1000 which are divisible by 13. a) 28405 b) 24805 c) 28540 d) 28450 Answer : (a) 28405 Hint : Sum of series 507, 520, 533.988
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