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How To Solve Arithmetic Progression Questions Quickly


How To Solve Arithmetic Progression Questions Quickly
Arithmetic progression is a sequence of terms such that the difference between the consecutive terms is a constant value . Or we can say that a succession of terms or numbers formed and arranged in a definite order according to a definite rule is called a Arithmetic Progression.
How to solve Arithmetic Progressions questions easily
- Definition
Arithmetic progression (AP) is a sequence of numbers in which each term after the first is obtained by adding a constant d to the preceding term. The constant d is called common difference.
An AP is represented in the form a, (a + d), (a + 2d), (a + 3d), …
where a = the first term, and d = the common difference.
General form, Tn = a+ (n-1)d
Type 1: Find nth term of an AP: tn = a + (n – 1)d
Question 1. If tn = 20, d = -1 and n = 16, then the first term is?
Options:
A. 36
B.-35
C. 35
D. 5
Solution: We know that, tn = a + (n – 1)d
20 = a + (16 – 1) * (-1)
20 = a + (15) * (-1)
20 = a – 15
-a = – 15 – 20
– a = – 35
a = 35
Correct option: C
Question 2. What is the position of 62 in the following series 2, 5, 8, ….?
Options:
A. 21
B. 22
C. 23
D. 20
Solution: In the given series,
a = 2
d = 3
tn = 62
tn = a + (n – 1)d
62 = 2 + (n -1) * 3
62 = 2 + 3n – 3
62 – 2 + 3 = 3n
63 = 3n
n = 21
Correct option: A
Question 3. Find the 11th term of the series 1, 3.5, 6, 8.5, ……
Options:
A. 25
B. 26
C. 23.5
D. 24.5
Solution: In the given series a = 1, d = 2.5
n =11
tn = a + (n – 1)d
tn = 1 + (11 – 1) * 2.5
tn = 1 + 10 * 2.5
tn = 1+ 25
tn = 26
Correct option: B
Type 2: Number of terms in the series: n = \left [ \frac{l – a}{d} \right ] + 1
Question 1.Find the number of terms in the series 2, 4, 6, . . .98
Options:
A. 50
B. 49
C.51
D. 48
Solution We know that, n = \left [ \frac{l – a}{d} \right ] + 1
a (first term) = 2
l (last term) = 98
d (common difference) = 2
n= [(98-2) / 2 ]+ 1
n = (96/2) + 1
n = 48 + 1
n = 49
Correct option: B
Question 2.Find the last number of the series where, a = 3, d = 5, and number of terms are 45
Options:
A. 225
B.223
C.230
D. 222
Solution According to the question,
a = 3, d = 5, n = 45
n = \left [ \frac{l – a}{d} \right ] + 1
45 = [(l-3)/5]+ 1
45-1 = (l-3)/5
44*5 = l – 3
220 = l-3
l=223
Correct option: B
Question 3.Find the total odd numbers in the series 3….. 99.
Options:
A. 49
B.51
C. 50
D. 60
Solution: The common difference is 2 because it is an odd number series.
We know that,
n = \left [ \frac{l – a}{d} \right ] + 1
a = 3
l = 99
d (common difference) = 2
n= (99-3)/2 + 1
n = 96/2 + 1
n = 48 + 1
n = 49
Correct option: A
Type 3: Sum of first ‘n’ terms of the series: S_{n} = \frac{n}{2}\left [ 2a + (n-1) d\right ] \text{ or } \frac{n}{2}(a + l)
Question 1.Sanjana has one paisa on 1st of January with her. She got 2 paisa on the 2nd January and 3 paisa on 3rd of January. In similar fashion, what will be her saving after 365 days?
Options:
A. 66700 paisa
B. 66795 paisa
C. 65795 paisa
D. 66790 paisa
Solution: According to the question, the series formed is 1,2,3,4,5,……
Therefore, a = 1 and d = 1
S_{n} = \frac{n}{2}\left [ 2a + (n-1) d\right ]
n = 365
S365 = 365/2*[2 * 1 + (365 − 1) * 1]
S365 = 66795 paisa
Correct option: C
Question 2.Find the sum of the series 62 + 60 + 58 …….. + 30
Options:
A. 728
B.780
C.719
D. 782
Solution: In the given series a = 62, d = -2, tn = 30
tn = a + (n – 1)d
30= 62 + (n – 1) * -2
30 = 62 – 2n + 2
2n = 62-30+2
34 = 2n
n = 17
S_{n} = \frac{n}{2}(a + l)
S17 = 17/2 (62 + 30)
S17 = 17/2(92)
S17 = 782
Correct option: D
Question 3.If the sum of 5 consecutive numbers is 100 then find out the first number.
Options:
A. 20
B. 25
C. 18
D. 16
Solution: It is clear from the question that d = 1 because there are 5 consecutive numbers
n = 5
Sn= 100
Let the first number ‘a’ be x
S_{n} = \frac{n}{2}\left [ 2a + (n-1) d\right ]
100 = 5/2 [2a + (5 − 1) * 1]
100 = 5/2[2a + (4) * 1]
100*2/5 = [2a + 4]
40 = 2a + 4
a = 18
So, the first number is 18 and the complete series is 18, 19, 20, 21, and 22.
Correct option: C
Type 4: Arithmetic mean of the series: b = \frac{(a + c)}{2}
Question 1.Find the arithmetic mean of first five prime numbers.
Options:
A. 4.5
B. 6.5
C. 5.6
D. 6.4
Solution: We know that, arithemetic mean of a,b,c is
b = (a + c)/2
Here, five prime numbers are 2, 3, 5, 7 and 11
Therefore, there arithmetic mean (AM) =(2+3+5+7+11)/5 = 5.6
Correct option: C
Question 2. Find the arithmetic mean of first 11 whole numbers.
Options:
A. 4
B. 6
C.5
D.6
Solution: We know that,
b = \frac{1}{2(a + c)}
Therefore, arithmetic mean (AM) = 0+1+2+3+4+5+6+7+8+9+10)/11= 5
Correct option: C
Question 3. Find the arithmetic mean of series 2, 4, 6 , 8….100
Options:
A. 51
B.50
C.25
D.49
Solution: In the given series, a = 2, d = 2, l = 100
To find the total number of term we can use the formula
n= [\frac{(l-a)}{d}] + 1
n = [(100-2)/2]+ 1
n = 50
Now, to find the sum of the series we can use the formula Sn = [2a + (n − 1) d] OR n/2 (a+l)
Sn = n/2 (a+l)
Sn = n/2(a+l)
S50 = 50/2(2+l00)
S50 = 25 (102)
S50 = 2550
Now, AM = 2550/50 = 51
Correct option: A
Read Also –Tips and Tricks to solve Boat and stream Problems
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