Answer Paper. Time: 2 Hrs Max.

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1 Page 1 of 14 Algebra - Unit Test 1 Arithmetic Progression and Geometric Progression Answer Paper Time: 2 Hrs Max. Marks: 40 Note 1). All questions are compulsory. Note 2). Use of calculator is not allowed. Q. 1. Solve the following: (Any 5 out of 6) (5 marks) i) The n th term of a sequence is 2n + 3. Find fourth term of the sequence. Here n th term i.e. t n = 2n + 3. Substituting n = 4, t 4 = = = 11 The fourth term of the sequence is 11. ii) Find the fifth term in the sequence 3, 8, 13, 18. The given sequence is 3, 8, 13, 18. Here d the difference between two consecutive terms is constant and d = 5. Hence the fifth term = = 23. The fifth term in the sequence 3, 8, 13, 18 is 23. iii) For an A.P., find S 12 if a = 1 and d = 2. Here a = 1 and d = 2. We have,

2 Page 2 of 14 S 12 = 144. iv) Write the first two terms of the following Arithmetic Progression where the common difference d and the first term a are given. a = 5, d = 0 Here a = 5 and d = 0. Since a = 5, t 1 = 5 t 2 = t 1 + d = = 5 The first two terms of the given A. P. are 5, 5. v) Find the value of a and d for the A. P. : 12, 16, 20, 24,... Here a = 12 and d = = 4 a = 12 and d = 4. vi) State if the following list of numbers is a Geometic Progression. 1, 3, 9, 27,... The given list of numbers is 1, 3, 9, 27,... The ratio of two consecutive terms is 3 which is constant hence the given list of numbers is in G.P. The given list of numbers is a G.P. Q. 2. Solve the following : (Any 4 out of 5) (8 marks) i) Find t n for the sequence, if it is an A.P. : 24, 21, 18, 15,... In the given sequence 24, 21, 18, 15,... t 1 = 24, t 2 = 21, t 3 = 18, t 4 = 15,... Let's find the difference between consecutive terms. t 2 t 1 = = 3 t 3 t 2 = = 3 t 4 t 3 = = 3

3 Page 3 of 14 From the above, The difference between the consecutive terms is constant. The given sequence is an A.P. with a = 24 and d = 3. Now, t n = a + (n 1)d = 24 + (n 1) ( 3) = 24 3n + 3 = 27 3n t n = 27-3n ii). For an A.P., if t 1 = 100, t n = 1000 and n = 10. find S n. For an A.P.it is given that, 1) t 1 = 100 2) n = 10, 3) t n = 1000 Now the sum of the n terms of an A.P. is represented by S n = 5500 iii) How many terms are there in the A.P. 201, 208, 215,..., 369? It is given that, 201, 208, 215,..., 369 is an A.P. For a given A.P., Let there be n terms. Now in the given A.P., t 1 = 201, t 2 = 208, t 3 = 215 and t n = 369. and a = 201 and d = t 2 t 1 = = 7 Now we have, t n = a + (n 1)d Substituting values, 369 = (n 1) = 7(n 1)

4 Page 4 of = 7n 7 7n = 175 n = 25 Number of terms in the given A.P. are 25. iv) Find the 7th term of the A.P. 6, 10, 14,... The given A.P. is 6, 10, 14,... Here a = 6, t 1 = 6, t 2 = 10, t 3 = 1,... Let's find the d, the common difference. d = t 2 t 1 = 10 6 = 4 From the above, Now, t n = a + (n 1)d t 7 = 6 + (7 1) (4) t 7 = 6 + (6 4) t 7 = t 7 = 30 t 7 = 30 v) If for an A.P., S 33 = 5544, a = 8, find d. It is given that for an A.P., 1) S 33 = ) a = 8

5 Page 5 of 14 d = 10 vi) Find t n for the sequence, if it is A.P. : 1, 5, 9, 13,... In the given sequence 1, 5, 9, 13,... t 1 = 1, t 2 = 5, t 3 = 9,t 4 = 13,... Let's find the difference between consecutive terms. t 2 t 1 = 5 1 = 4 t 3 t 2 = 9 5 = 4 t 4 t 3 = 13 9 = 4 From the above, The difference between the consecutive terms is constant. The given sequence is an A.P. with a = 1 and d = 4. Now, t n = a + (n 1)d = 1 + (n 1) 4 = 1 + 4n 4 = 4n 3 t n = 4n 3 Q. 3. Solve the following : (Any 3 out of 5) (9 marks) i) First and last terms of an A.P. are 4 and 37 respectively. If common difference is 3, find sum of all the terms and number of terms.

6 Page 6 of 14 It is given that for an A.P., 1) a = 4, t n = 37 and d = 3 Now, t n = a + (n 1)d Using (1), 37 = 4 + (n 1)3 37 = 4 + 3n 3 37 = 1 + 3n 36 = 3n n = 12...(i) Number of terms are 12. Their sum is 246. ii) The sum of the first 55 terms of an A.P. is 3300, Find the 28 th term. It is given that for an A.P., S 55 = 3300 We have, Now, t n = a + (n - 1)d t 28 = a + (28-1)d t 28 = a + 27d...(ii) From (i) and (ii), t 28 = 60

7 Page 7 of 14 The 28 th term, t 28 is 60. iii) Consider the sequence 50, 75, 100,... Identify if this is an A.P. If so, find common difference and next two terms of the sequence. In the given sequence 50, 75, 100, t 1 = 50, t 2 = 75, t 3 = 100,... Let's find the difference between consecutive terms. t 2 t 1 = 75 ( 50) = 25 t 3 t 2 = 100 ( 75) = 25 From the above, The difference between the consecutive terms is constant. The given sequence is an A.P. with a = 50 and d = 25. We have find next two terms for a given A.P. The next two terms are t 4 and t 5 Now, t n = a + (n 1)d t 4 = 50 + (4 1) ( 25) t 4 = = 125 and t 5 = 50 + (5 1) ( 25) t 5 = = 150 1) Comman difference is -25 2) Next two terms are -125 and -150 iv) If the third and the sixth terms of an A.P. are 7 and 13 respectively, find a, d and write the A.P. For the given A.P., it is given that, 1) t 3 = 7 2) t 6 = 13 For an A.P. let a be the first term and d be the common difference of the A.P. t n = a + (n 1)d...(i) Using (i), t 3 = a + (3-1)d t 3 = a + 2d Using (1), 7 = a + 2d a + 2d = 7...(ii)

8 Page 8 of 14 Similarly using (2) and (i), t 6 = 13 = a + (6-1)d i.e. a + 5d = 13...(iii) Subtracting (ii) from (iii), a + 5d - a - 2d = d = 6 d = 2 Substituting value of d in (ii), a + 2(2) = 7 a + 4 = 7 a = 7-4 a = 3 Hence the A.P. is 3, 5, 7, 9, 11,... a, first term of the given A.P. is 3. d, common difference is 2. The A.P. is 3, 5, 7, 9, 11,... v) Which term of the A.P. 8, 11, 14, 17,... is 758? It is given that, 1) 8, 11, 14, 17,... is an A.P For an A.P. let a be the first term and d be the common difference of the A.P. t n = a + (n 1)d...(i) Using (1), a = 8...(ii) Now for an A.P. the difference between consecutive terms (d) is constant. Consider first and second terms of the given A.P. using (1), d = 11 8 = 3...(iii) We have to find which term of the given A.P. is 758 Let 758 be the n th term of an A.P. Using (i), 758 = a + (n 1)d Substituting values of a and d using (ii) and (iii), 758 = 8 + (n 1)3 i.e. 758 = 8 + 3n 3

9 Page 9 of = 5 + 3n 3n = 753 n = is 251 th term of the gven A.P. Q. 4. Solve the following : (Any 2 out of 3) (8 marks) i) Find how many 2 digit natural numbers are divisible by 4 and sum of all such numbers. We have to find 2 digit natural numbers that are divisible by 4. Since the numbers should be divisible by 4, the difference between two consecutive such numbers will be 4 and all these numbers will form an A.P. Two digit natural numbers start from 10 and end with 99. Now numbers between 10 and 99 that are divisible by 4 are 12, 16, 20,..., 96. This forms an A.P. with a = 12, d = 4 and t n = 96 Now t n = a + (n 1)d Substituting values of a, d and t n, 96 = 12 + (n 1) 4 96 = n 4 96 = 8 + 4n 4n = 96 8 = 88 n = 22 1) Total two digit natural numbers that are divisible by four are 22. 2) The sum of all such numbers is ii) The sequence 3, 5, 7,... forms an A.P. How many terms in this A.P. should be taken so that their sum is 120? It is given that, 1) 3, 5, 7,... forms an A.P.

10 Page 10 of 14 Here a = 3 t 1 = 3, t 2 = 5, t 3 = 7 Now d = t 2 t 1 = 5 3 = 2 Now we have to find n such that S n = 120 Since number of terms can not be negative, n = terms should be taken so that their sum is 120 iii) A shopkeeper's sales of goods decreased by Rs. 100 everyday in month of Dec. If sales on Dec 15 were Rs. 3000, find out the total sales of good in month of Dec and sales on Dec 1. It is given that, The shopkeeper's sales of goods decreased by Rs. 100 everyday in month of Dec Since sales on every day is less than that of previous day by Rs. 100, the difference between sales on two consecutive days will be Rs. 100 and all these numbers will form an A.P. with d = 100 Now sales on Dec 15 = Rs i.e. t 15 = 3000 Now t n = a + (n 1)d Substituting values of d and n, 3000 = a + (15 1)( 100) 3000 = a 1400 a = = Rs. 4400

11 Page 11 of 14 Total sales of goods in the month of Dec is Rs Sales on Dec 1. i.e. a is Rs Q. 5. Solve the following : (Any 2 out of 3) (10 marks) i) If 3rd term of an A.P. is 25 and 17th term exceeds 5th term by 120, find 29th term of the A.P. For an A.P. let a be the first term and d be the common difference of the A.P. t n = a + (n 1)d...(1) It is given that, 3rd term, t 3 = 25...(2) 17th term exceeds 5th term by (3) Using (1), 3rd term, t 3 = a + (3 1)d i.e. t 3= a + 2d Using (2), a + 2d = 25...(4) And 5th term, t 5 = a + (5 1)d i.e. t 5= a + 4d and 17 th term, t 17 = a + (17 1)d i.e. t 17 = a + 16d

12 Page 12 of 14 using (3), t 17 = t a + 16d = a + 4d d 4d = d = 120 d = 10 Now substituting value of d in (4), a + 2(10) = 25 a + 20 = 25 a = = 5 Substituting values of a, d and n = 29 in (1), t 29 = 5 + (29 1)(10) t 29 = 5 + (28)(10) t 29 = t 29 = th term of the A.P. is 285. ii) Babubhai borrows Rs and agrees to repay with a total interest of Rs.500 in 10 instalments, each instalment being less than the preceding instalment by Rs.10. What should be the first and the last instalment? It is given that, (1) Babubhai barrows Rs (2) The intersest paid = Rs.500 for total 10 instalmants. (3) Every instalment is less than the previous by Rs.10 The sequence of insatlments forms an A.P. with d = 10 The total amount paid back = amount borrowed + interest paid The total amount paid back = = 4500 Thus S n = 4500, n = 10 We have, We also have,

13 Page 13 of 14 The first and the last instalments are Rs. 495 and Rs. 405 respectively. iii) Consider two G.P. as, a, am, am 2, am 3,... and am, a 2 m, a 3 m,... If the third term of first G.P. is twice the third term of the second G.P. and fourth term of first G.P. is thrice the fourth term of second G.P. then find the first three terms of each G.P. The given G.P. are, a, am, am 2, am 3,... am, a 2 m, a 3 m, a 4 m,... It is given that third term of first G.P. is twice that of corresponding term of second G.P. am 2 = 2a 3 m m = 2a 2...(1) And fourth term of first G.P. is thrice that of fourth term of second G.P. am 3 = 3a 4 m m 2 = 3a 3...(2) Substituting value of m from (1) in (2), (2a 2 ) 2 = 3a 3

14 Page 14 of 14 4a 4 = 3a 3