Arithmetic Progression worksheet-1

~Anshul Gurjar


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1. In the Arithmetic process, if a = 28, d = -4, n = 7, then:


(a) 4


(b) 5


(c) 3


(d) 7


Answer: a


Description: By AP,


a = a + (n-1) d


= 28+ (7-1) (- 4)


= 28 + 6 (-4)


= 28-24


i = 4


2. If = 10 and d = 10, the first four goals will be:


(a) 10,30,50,60


(b) 10,20,30,40


(c) 10,15,20,25


(d) 10,18,20,30


Answer: b


Definition: a = 10, d = 10


a1 = a = 10


a2 = a1 + d = 10 + 10 = 20


a3 = a2 + d = 20 + 10 = 30


a4 = a3 + d = 30 + 10 = 40


3.The first sentence and the standard variation of A.P. 3,1, -1, -3 states:


(a) 1 and 3


(b) -1 and 3


(c) 3 and 2


(d) 2 and 3


Answer: c


Definition: First term, a = 3


Standard difference, d = Second term - First term


⇒ 1 - 3 = 2


⇒ d = 2


4.30th Time A.P: 10.7, 4,…,


(a) 97


(b) 77


(c) -77


(d) -87


Answer: c


Definition: Given,


A.P. = 10, 7, 4,…


First term, a = 10


Normal difference, d = a2 - a1 = 7−10 = −3


As we know, by A.P.


i = a + (n - 1) d


Pricing;


a30 = 10+ (30−1) (- 3)


a30 = 10+ (29) (- 3)


a30 = 10−87 = −77


5.11 time of A.P. -3, -1 / 2 ,, 2…. Is


(a) 28


(b) 22


(c) -38


(d) -48


Answer: b


Description: A.P. = -3, -1 / 2 ,, 2…


First term a = - 3


Standard difference, d = a2 - a1 = (-1 / 2) - (- 3)


⇒ (-1 / 2) + 3 = 5/2


Nth term;


i = a + (n - 1) d


Pricing;


a11 = 3+ (11-1) (5/2)


a11 = 3+ (10) (5/2)


a11 = -3 + 25


a11 = 22


6. The missing word in the AP: __, 13, __, 3 is:


(a) 11 and 9


(b) 17 and 9


(c) 18 and 8


(d) 18 and 9


Answer: (c)


Definition: a2 = 13 no


a4 = 3


Nth time for AP;


i = a + (n - 1) d


a2 = a + (2-1) d


13 = a + d ………………………. (i)


a4 = a + (4-1) d


3 = a + 3d ………… .. (ii)


Subtracting equation (i) from (ii), we find,


- 10 = 2d


d = - 5


Now enter the value of d in 1


13 = a + (- 5)


a = 18 (first name)


a3 = 18+ (3-1) (- 5)


= 18 + 2 (-5) = 18-10 = 8 (third term).


7.How long for A.P. 3, 8, 13, 18,… is 78?


(a) 12


(b) 13


(c) 15


(d) 16


Answer: (d)


Description: Given, 3, 8, 13, 18,… by AP.


First term, a = 3


Normal difference, d = a2 - a1 = 8 - 3 = 5


Enable nth name for A.P. given 78. Now as we know,


i = a + (n - 1) d


Therefore,


78 = 3+ (n −1) 5


75 = (n - 1) 5


(n - 1) = 15


n = 16


8. The 21st name of the AP is its first 3 and 4 names are:


(a) 17


(b) 137


(c) 143


(d) -143


Answer: b


Definition: First term = -3 and second term = 4


a = -3


d = 4-a = 4 - (- 3) = 7


a21 = a + (21-1) d


= -3 + (20) 7


= -3 + 140


= 137


9. If the 17th period of A.P. exceeds its 10th time by 7. The most common difference is:


(a) 1


(b) 2


(c) 3


(d) 4


Answer: (a)


Definition: Nth time in AP by:


a = a + (n-1) d


a17 = a + (17−1) d


a17 = a + 16d


Vice versa,


a10 = a + 9d


Given,


a17 - a10 = 7


Therefore,


(a + 168d) - (a + 9d) = 7


7d = 7


d = 1


Therefore, the common difference is 1.


10. The number of 4 times between 10 and 250 is:


(a) 50


(b) 40


(c) 60


(d) 30


Answer: (c)


Definition: 4 repetitions after 10 are:


12, 16, 20, 24,…


So here, a = 12 and d = 4


Now, 250/4 gives the remaining 2. So, 250 - 2 - 248 is from 2.


12, 16, 20, 24,…, 248


Therefore, the term nth, i = 248


As we know,


i = a + (n - 1) d


248 = 12+ (n-1) × 4


236/4 = n-1


59 = n-1


n = 60


11. Term 20 is from the last term of A.P. 3, 8, 13,…, 253 are:


(a) 147


(b) 151


(c) 154


(d) 158


Answer: (d)


Description: Given, A.P. by 3, 8, 13,…, 253


Standard difference, d = 5.


By postponing,


253, 248, 243,…, 13, 8, 5


Therefore,


a = 253


d = 248 - 253 = −5


n = 20


With nth formula,


a20 = a + (20−1) d


a20 = 253+ (19) (- 5)


a20 = 253−95


a20 = 158


12. The total of the first five dimensions of 3 is:


(a) 45


(b) 55


(c) 65


(d) 75


Answer: (a)


Definition: The first five duplicates are 3,6,9,12 and 15


a = 3 no d = 3


n = 5


ISum, Sn = n / 2 [2a + (n-1) d]


S5 = 5/2 [2.3+ (5-1) 3]


= 5/2 [6 + 12]


= 5/2 [18]


= 5 x 9


= 45

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