ID : ww-0-arithmetic-progressions [] Grade 0 Arithmetic Progressions For more such worksheets visit www.edugain.com 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 - 435. What is the sum of the f irst 40 terms of this AP? a. -3665 b. -3650 c. -3660 d. -3670 (0) What is the middle term of the arithmetic progression 8, 4, 20,...,332? a. 76 b. 82 c. 70 d. 64
() 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. -324 b. -334 c. -304 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. -20 206 Edugain (www.edugain.com). All Rights Reserved Many more such worksheets can be generated at www.edugain.com
Answers ID : ww-0-arithmetic-progressions [3] () -2575 According to the question, a n = 4-9n Theref ore, a = 5, a 2 = -4, a 3 = -3 Now, the required AP: 5, -4, -3... 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 = -2575 Thus, the sum of the f irst 25 terms of this AP is -2575. (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 - 660 4 a = 2260 + 660 4 a = 2920 a = 2920/4 a = $730
(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 = 52 ------() 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 = 6 + 78(8) = 630
(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)] -3300 = [-6n 2 + (8n)] 6n 2-8n - 3300 = 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.
(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 = 0 + 3 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
(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 [80 + 300] 2 S 23 = 4370 Thus, the sum of all natural numbers between 80 and 300 which are divisible by 0 is 4370. (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) 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) c. -3660 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 = -65... () It is also given that sum of f irst 5 terms is -435, S 5 = -435 (5/2)[2a + (5 - ) d] = -435 5a + 5(5-)(d/2) = -435 5a + 05d = -435... (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
(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 = [ 8 + 332 ]/2 x = 70
() 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 = 3 056 -----(2) By putting the value of 'd' in equation (), we get: a = 6-2 3 056 = 3 056 Step 5 Now, a 352 = a + (352 - )d = 3 056 + 35 3 056 =
= 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 = -424 + (2 - )(0) T 2 = -424 + (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