Grade 10 Arithmetic Progressions

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1 ID : us-0-arithmetic-progressions [] Grade 0 Arithmetic Progressions For more such worksheets visit Answer t he quest ions () The sum of f irst 9 terms of an arithmetic progression is -234 and sum of its f irst 20 terms is What is the sum of the f irst 45 terms of this AP? (2) Find number of terms in the arithmetic progression 0, 7, 24, 3,... such that sum of the series is 455? (3) The nth term of an arithmetic progression is given by the equation n. What is the sum of the f irst 30 terms of this AP? (4) What is the value of the f ollowing series: ? (5) The sum of f irst 5 terms of an arithmetic progression is 65. The sum of the f irst 0 terms is 5. What is the sum of the f irst 5 terms of the AP? (6) The sum of the f irst three terms of an AP is 2, and their product is 504. What is the value of the 3th term? (7) The sum of the 7 th element and the 5 th element of an arithmetic progression is 98. The sum of the 0 th and the 8 th element is 2. What is the value of the 33 rd term? Choose correct answer(s) f rom given choice (8) If the sum of the f irst n terms of an arithmetic progression is given by the equation 24n + 4n 2, then what is the 7 th element of this AP? a. 56 b. 64 c. 48 d. 40 (9) Choose the arithmetic progression f rom the list below: a. 6, -8, -22, -36 b. 7, -8, -2, -36 c. 6, -8, -2, -36 d. 6, -22, -36, -50 (0) If the 85 th element of an arithmetic progression is 97 and the 50 th element is 707, then what is the 5 th element? a. 77 b. 49 c. 63 d. 9 () What is the sum of all three-digit numbers that are divisible by 23? a. 27 b. 28 c. 3 d. 30

2 (2) In an arithmetic progression, the value of the 3 th element is ID : us-0-arithmetic-progressions [2], and the value of th element is 3. What is the value of 3 th element? a. 2 b. 0 c. d. 3 (3) A sequence of numbers that are in arithmetic progression starts with 7 and ends with 224. What is the sum of the th term f rom the beginning of the sequence and the th term f rom the end of the sequence? a. 239 b. 243 c. 24 d. 240 (4) In an arithmetic progression, if the value of the ath term is b, and the value of the bth term is a, then what is the value of the nth term? a. a - b - n b. n - b - a c. a + b - n d. a - b + n (5) Which term in the arithmetic progression 7, 2, 7, 22 will be 35 more than the 20 th term? a. 46 b. 49 c. 47 d Edugain ( All Rights Reserved Many more such worksheets can be generated at

3 Answers ID : us-0-arithmetic-progressions [3] () Step Sum of f irst n terms of arithmetic progression, S n = (n/2)[2a + (n - ) d] It is given that sum of f irst 9 terms is -234, S 9 = -234 (9/2)[2a + (9 - ) d] = a + 9(9-)(d/2) = a + 36d = () It is also given that sum of f irst 20 terms is -400, S 20 = -400 (20/2)[2a + (20 - ) d] = a + 20(20-)(d/2) = a + 90d = (2) On solving Eq. () and (2), we get, a = 6 and d = -8 Now, sum of f irst 45 terms, S 45 = (45/2)[2a + (45 - ) d] S 45 = (45/2)[2(6) + (45 - ) (-8)] S 45 = (45/2)[-340] S 45 = -7650

4 (2) 35 ID : us-0-arithmetic-progressions [4] Step The given AP is, 0, 7, 24, 3,... Now, f irst term a = 0, Dif f erence between successive terms d = 7-0 = 7 We know that, S n = n 2 [2a + (n - )d] Let's assume, the given AP have 'n' number of terms. According to the question, S n = 455 Theref ore, 455 = n 2 [(2 0) + (n - )(7)] 455 = n 2 [7n + (3)] 9030 = [7n 2 + (3n)] 7n 2 + 3n = 0 The number of terms neither negative nor in f raction, the value of 'n' must be positive. By solving above quadratic equation, we get: n = 35 Theref ore, this AP should have 35 terms.

5 (3) 4 ID : us-0-arithmetic-progressions [5] Step According to the question, a n = n Theref ore, a = 7, a 2 = 6, a 3 = Now, the required AP: 7, 6,... d = a 2 - a = 6-7 = 9 The sum of the f irst 30 terms of this AP(S 30 ) = 30 2 [2a + (30 - )d] 30 2 [(2 7) + (30 - )(9)] = 30 [(2 7) + (29) (9)] 2 = 4 Thus, the sum of the f irst 30 terms of this AP is 4.

6 (4) 848 ID : us-0-arithmetic-progressions [6] Step From f irst f our terms, we can observe that this is an arithmetic progression with dif f erence of 6 If there are n terms in the arithmetic progression, its last term is given by, T n = T + (n-)d 98 = T + (n-)d 98 = 8 + (n-)6 n- = (98-8)/6 n- = 5 n = 6 Now sum of arithmetic progression can be f ound using standard f ormula, S n = (n/2)[2t + (n-)d ] S 6 = (6/2)[2(8) + (6-)(6) ] S 6 = (6/2)[ 06 ] S 6 = 848

7 (5) 570 ID : us-0-arithmetic-progressions [7] Step Sum of f irst n terms of arithmetic progression, S n = (n/2)[2a + (n - ) d] It is given that sum of f irst 5 terms is 65, S 5 = 65 (5/2)[2a + (5 - ) d] = 65 5a + 5(5-)(d/2) = 65 5a + 0d = () It is also given that sum of f irst 0 terms is 5, S 0 = 5 (0/2)[2a + (0 - ) d] = 5 0a + 0(0-)(d/2) = 5 0a + 45d = 5... (2) On solving Eq. () and (2), we get, a = 3 and d = 5 Now, sum of f irst 5 terms, S 5 = (5/2)[2a + (5 - ) d] S 5 = (5/2)[2(3) + (5 - ) (5)] S 5 = (5/2)[76] S 5 = 570 (6) -5

8 (7) 297 ID : us-0-arithmetic-progressions [8] Step If d is the dif f erence between consecutive terms, the n th term of arithmetic progression is, T n = T + (n-) d It is given that sum of the 7 th element and the 5 th element is 98 T 7 + T 5 = 98 T + (7-) d + T + (5-) d = 98 2T + 20d = 98...() It is also given that sum of the 0 th element and the 8 th element is 2 T 0 + T 8 = 2 T + (0-) d + T + (8-) d = 2 2T + 26d = 2...(2) On subtracting Eq. () f rom (2) [2T + 26d] - [2T + 20d] = 2-98 (26-20) d = 54 (6)d = 54 d = 9 On subtracting d in Eq. () 2T + 20(9) = 98 2T = (98) - (80) T = [(98) - (80)]/2 = 9 Step 6 Theref ore 33 rd term, T 33 = T + (33-) d T 33 = 9 + (33-)(9) T 33 = 9 + (288) = 297

9 (8) a. 56 ID : us-0-arithmetic-progressions [9] Step For an arithmetic progression with f irst term = a, and common dif f erence of successive members = d, sum of f irst n terms is given by, S n = (n/2)[2a + (n-)d ] It is given that sum of f irst n terms is 24n + 4n 2, S n = 24n + 4n 2 (n/2)[2a + (n-)d ] = 24n + 4n 2 (n/2)[2a + nd - d ] = 24n + 4n 2 na + (d/2)n 2 - (d/2)n = 24n + 4n 2 (a - d/2)n + (d/2)n 2 = 24n + 4n 2... () On comparing f actors of n 2 in Eq. () d/2 = 4 d = 8 On comparing f actors of n in Eq. () a - d/2 = 24 a - d/2 = 24 a - 4 = 24 a = a = 28 Now we can f ind 7 th term using standard f ormula f or arithmetic progression, a n = a + (n-) d a 7 = 28 + (7-) 8 a 7 = a 7 = 56

10 (9) a. 6, -8, -22, -36 ID : us-0-arithmetic-progressions [0] Step In an arithmetic progression, dif f erence between consecutive terms is f ixed. Lets check this f or all the given progressions Dif f erences in series 6, -8, -2, -36 are (-8) - (6) = -4 (-2) - (-8) = -3 (-36) - (-2) = -5 Dif f erences in series 6, -22, -36, -50 are (-22) - (6) = -28 (-36) - (-22) = -4 (-50) - (-36) = -4 Dif f erences in series 6, -8, -22, -36 are (-8) - (6) = -4 (-22) - (-8) = -4 (-36) - (-22) = -4 Dif f erences in series 7, -8, -2, -36 are (-8) - (7) = -5 (-2) - (-8) = -3 (-36) - (-2) = -5 Step 6 It can be seen that dif f erence is f ixed only f or series (6, -8, -22, -36), theref ore (6, -8, -22, -36) is an arithmetic progression

11 (0) a. 77 ID : us-0-arithmetic-progressions [] Step If d is the dif f erence between consecutive terms, the n th term of arithmetic progression is, T n = T + (n-) d It is given that 85 th element is 97, T + (85-) d = 97 T + 84d = () Similarly it is given that 50 th element is 707, T + (50-) d = 707 T + 49d = (2) On subtracting equation () f rom (2), 84d - 49d = (97) - (707) 35d = 490 d = (490)/(35) d = 4 Replace d by 4 in equation (2) T + 49(4) = 707 T = (4) T = 2 Step 6 Theref ore 5 th term will be, T 5 = T + (5-) d T 5 = 2 + (5-) 4 T 5 = 77

12 () b. 28 ID : us-0-arithmetic-progressions [2] Step All numbers which are divisible by 23, f orms an arithmetic progression with dif f erences between consecutive terms to be 23 We know smallest three digit number is 00. On dividing 00 by 23 we get remainder of 8 Theref ore if we if we add remaining (23-8 = 5) to 00, resultant number ( = 5) will be f ully divisible by 23 Theref ore f irst number of the arithmetic progression is 5 Similarly, largest three digit number is 999. On dividing 999 by 23 we get remainder of 0 Theref ore if we if we subtract 0 f rom 999, resultant number (999-0 = 989) will be f ully divisible by 23 Theref ore last number of the arithmetic progression is 989 If there are total N terms in series, N th term is given by T N = T + (N-)d 989 = 5 + (N-)(23) 23(N - ) = (N - ) = 874 N - = 874/23 N - = 38 N = 38 + N = 39 Now sum of arithmetic progression can be f ound using standard f ormula, S N = (N/2)[T + (N-)d] S N = (39/2)[2 5 + (39-)(23)] S N = (39/2)[ ] S N = (39/2)[04] S N = 28

13 (2) c. ID : us-0-arithmetic-progressions [3] Step We know that, a n = a + (n - )d, Where, n = number of elements in an AP(arithmetic progression), a = f irst element of an AP, d = common dif f erence. According to the question, a 3 = a 3 = a + (3 - )d, a + (3 - )d = a + 2d = a = - 2d -----() a = 3, a = a + ( - )d By putting the value of 'd' f rom equation (), we get: - 2d + 24d = 3 2d = 3 - d = (2) By putting the value of 'd' in equation (), we get: a = = Now, a 3 = a + (3 - )d = =

14 = ID : us-0-arithmetic-progressions [4] Step 6 Thus, the value of 3 th element is. (3) c. 24

15 (5) c. 47 ID : us-0-arithmetic-progressions [5] Step If you inspect this series, you will f ind that dif f erence between consecutive terms is 5 Since dif f erence between consecutive terms (i.e. distance of ) is 5, dif f erence between terms with distance n will be 5n. i.e. T a - T b = 5(a-b) Lets required term be m th term, T m - T 20 = 35 (m - 20)5 = 35 m - 20 = 35/5 m - 20 = 27 m = m = 47