Grade 10 Arithmetic Progressions

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1 ID : ww-0-arithmetic-progressions [] Grade 0 Arithmetic Progressions For more such worksheets visit Answer t he quest ions () The nth term of an arithmetic progression is given by the equation 4-9n. What is the sum of the f irst 25 terms of this AP? (2) A sum of $ 2260 is set aside f or giving 4 prizes in a competition. If each prize is $ 0 lower than the preceding prize, what is the amount given f or the f irst prize? (3) The sum of f irst 6 terms of an arithmetic progression is 56. The ratio of term 5 th to term 25 th of the AP is 59:99. What is the value of the 79 th term of the AP? (4) Find number of terms in the arithmetic progression 6, 0, -6, -2,... such that sum of the series is -650? (5) If the nth term of an arithmetic progression is given by 0 + 3n, then what is the sum of the f irst 45 terms of the AP? (6) What is the sum of all natural numbers between 80 and 300 which are divisible by 0? Choose correct answer(s) f rom given choice (7) What is the dif f erence between the 34 th element and 7 th element in the arithmetic progression 0,, -8, -7? a. -62 b. -53 c. -7 d. -44 (8) The th term of an arithmetic progression has the value 0. Which term of the AP is 2 times the 9 th element? a. 27 b. 29 c. 30 d. 25 (9) The sum of f irst 0 terms of an arithmetic progression is -65 and sum of its f irst 5 terms is What is the sum of the f irst 40 terms of this AP? a b c d (0) What is the middle term of the arithmetic progression 8, 4, 20,...,332? a. 76 b. 82 c. 70 d. 64

2 () In an arithmetic progression, the value of the 22 nd element is 6 ID : ww-0-arithmetic-progressions [2], and the value of 6 th element is 22. What is the value of 352 nd element? a. 2 b. 0 c. d. - (2) Which is the 2 th term f rom the end of the arithmetic progression 6, -4, -4,...,-424? a b c d. -34 (3) A sequence of numbers that are in arithmetic progression starts with 9 and ends with 26. What is the sum of the 9th term f rom the beginning of the sequence and the 9th term f rom the end of the sequence? a. 238 b. 233 c. 234 d. 235 (4) How many elements are there in the arithmetic progression 0,, -8,..., -88? a. 26 b. 2 c. 23 d. 22 (5) The sum of the f irst three terms of an AP is 2, and their product is 280. What is the value of the 3th term? a. -29 b. -32 c. -26 d Edugain ( All Rights Reserved Many more such worksheets can be generated at

3 Answers ID : ww-0-arithmetic-progressions [3] () According to the question, a n = 4-9n Theref ore, a = 5, a 2 = -4, a 3 = -3 Now, the required AP: 5, -4, d = a 2 - a = -4-5 = -9 The sum of the f irst 25 terms of this AP(S 25 ) = 25 2 [2a + (25 - )d] 25 2 [(2 5) + (25 - )(-9)] = 25 [(2 5) + (24) (-9)] 2 = Thus, the sum of the f irst 25 terms of this AP is (2) $ 730 Since dif f erence between consecutive prize amount it f ixed ($0), these prize amounts f orm an arithmetic progression with dif f erence of -$0 Sum of f irst n terms of arithmetic progression, S n = (n/2)[2a + (n - ) d] S 4 = (4/2)[2a + (4 - ) (-0)] 2260 = (4/2)[2a + (4 - ) (-0)] 2260 = 4 a + (4/2)(4 - ) (-0) 2260 = 4 a a = a = 2920 a = 2920/4 a = $730

4 (3) 630 ID : ww-0-arithmetic-progressions [4] Sum of the f irst n terms of arithmetic progression, Sn = (n/2)[2a + (n - )d]. n th term of arithmetic progression, a n = a + (n - )d]. It is given that sum of the f irst 6 terms is 56, S 6 = 56 (6/2)[2a + (6 - )d] = 56 (6/2)[2a + 5d] = 56 2a + 30d = 32 2a + 5d = () Now 5 th and 25 th terms can be f ound as f ollowing, a 5 = [a + (5 - )d] = a + 4d, a 25 = [a + (25 - )d] = a + 24d It is also given that the ratio of term 5 th to term 25 th of the AP is 59:99, a + 4d = 59 a + 24d 99 99a + 386d = 59a + 46d 99a - 59a = 46d -386d 40a = 30d a = (3/4) d (2) On substiuting value of a f rom equation () in equation (2), we get, d = 8 Step 5 Now, a = (3/4) d a = (3/4) (8) a = 6 Step 6 Now, the value of the 79 th term of the AP, a 79 = a + (79 - )d = (8) = 630

5 (4) 25 ID : ww-0-arithmetic-progressions [5] The given AP is, 6, 0, -6, -2,... Now, f irst term a = 6, Dif f erence between successive terms d = 0-6 = -6 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 = -650 Theref ore, -650 = n 2 [(2 6) + (n - )(-6)] -650 = n 2 [-6n + (8)] = [-6n 2 + (8n)] 6n 2-8n = 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 = 25 Theref ore, this AP should have 25 terms.

6 (5) 3555 ID : ww-0-arithmetic-progressions [6] 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, T n = 0 + 3n T + (n-) d = 0 + 3n (T -d ) + d(n) = 0 + 3n On comparing the terms in above equation, we get d = 3 and (T - d) = 0 T = 0 + d T = T = 3 Now sum of f irst 45 terms can be calculated using standard f ormula, S n = (n/2)[2t + (n-)d ] S 45 = (45/2)[2(3) + (45-)(3)] S 45 = (45/2)(58) S 45 = 3555

7 (6) 4370 ID : ww-0-arithmetic-progressions [7] The all natural numbers between 80 and 300, which are divisible by 0 are, 80, 90, 00,..., 300 If we look at these numbers, we notice that the all terms have same common dif f erence and hence the series is in AP. First term T = 80 Last tern T n = 300 Dif f erence between successive numbers d = 0 Number of terms n = + (T n - T ) d n = + n = 23 (300-80) 0 Now, Sum S n = n 2 (T + T n ) S 23 = 23 [ ] 2 S 23 = 4370 Thus, the sum of all natural numbers between 80 and 300 which are divisible by 0 is (7) b. -53 If you inspect this series, you will f ind that dif f erence between consecutive terms is -9 The n th term of this arithmetic progression will be, T n = T + (n-) d Theref ore dif f erence between n th and m th term will be, T n - T m = [T + (n-) d] - [[T + (m-) d]] T n - T m = [(n-m) d] T 34 - T 7 = (34-7)(-9) T 34 - T 7 = -53

8 (8) a. 27 ID : ww-0-arithmetic-progressions [8] 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 than th term is 0, T + (-) d = 0 T = -0 d...() Let m th term is 2 times the 9 th term, T m = 2 [T 9 ] T + (m-) d = 2 [T + (9-) d] T + (m-) d = 2 T + 36 d (m-) d = T + 36 d...(2) On replacing value of T f rom Eq. in Eq. (2), (m-) d = (-0 d) + 36 d (m-) = ()(-0) + 36 (m-) = 26 m = 27

9 (9) c ID : ww-0-arithmetic-progressions [9] Sum of f irst n terms of arithmetic progression, S n = (n/2)[2a + (n - ) d] It is given that sum of f irst 0 terms is -65, S 0 = -65 (0/2)[2a + (0 - ) d] = -65 0a + 0(0-)(d/2) = -65 0a + 45d = () It is also given that sum of f irst 5 terms is -435, S 5 = -435 (5/2)[2a + (5 - ) d] = a + 5(5-)(d/2) = a + 05d = (2) On solving Eq. () and (2), we get, a = 6 and d = -5 Step 5 Now, sum of f irst 40 terms, S 40 = (40/2)[2a + (40 - ) d] S 40 = (40/2)[2(6) + (40 - ) (-5)] S 40 = (40/2)[-83] S 40 = -3660

10 (0) c. 70 ID : ww-0-arithmetic-progressions [0] If n th term is the middle term, there will be 2n+ terms in the series The dif f erence between n th term and f irst term will be same as the dif f erence between last terms and n th term So if n th term is x, f irst term will be x - δ and last term will be x + δ We can see that x is middle point (or average) of f irst and last term. i.e., x = [( x - δ) + (x + δ) ]/2 x = [ ]/2 x = 70

11 () c. ID : ww-0-arithmetic-progressions [] 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 22 = a 22 = a + (22 - )d 6, a + (22 - )d = 6 a + 2d = 6 a = 6-2d -----() a 6 = 22, a 6 = a + (6 - )d By putting the value of 'd' f rom equation (), we get: 6-2d + 5d = 22-6d = 22-6 d = (2) By putting the value of 'd' in equation (), we get: a = = Step 5 Now, a 352 = a + (352 - )d = =

12 = ID : ww-0-arithmetic-progressions [2] Step 6 Thus, the value of 352 nd element is. (2) d. -34 Given arithmetic progression is 6, -4, -4,...,-424 If we reverse order of arithmetic progression, the resultant series is also an arithmetic progression. Theref ore -424,..., -4, -4, 6 is also an arithmetic progression with dif f erence d = 0 The n th term if arithmetic progression is, T n = T + (n - )d Theref ore 2 th term of reverse arithmetic progression will be, T 2 = (2 - )(0) T 2 = (0) T 2 = -34 (3) d. 235 (4) c. 23 If you inspect this series, you will f ind that dif f erence between consecutive terms is -9 First term of the series is 0. If there are n terms in this arithmetic progress, last term will be, T n = T + (n-) d -88 = 0 + (n-)(-9) (n-) = (-88-0)/(-9) (n-) = 22 n = 23