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Arithmetic Progressions Shortcut, tricks, and tips
Arithmetic Progression tricks, shortcuts, and tips
Arithmetic Progression can applied in real life by analyzing a certain pattern that we see in our daily life. In this page tips and tricks for Arithmetic Progression as well as different types of questions and their answers is given for solving questions of arithmetic progression in less time.
Tips and Tricks for Arithmetic Progression To Solve Arithmetic Progression in Aptitude
- An Arithmetic Progression or Arithmetic Sequence is a sequence of numbers or terms in a way that the difference between the consecutive terms is constant. Here, are some easy tips and tricks for you solve Arithmetic Progression questions quickly, easily, and efficiently in competitive exams.
An Arithmetic Progression is Represented in the form a, (a + d), (a + 2d), (a + 3d), …
where a = the first term, and d = the common difference.
n is number of Terms
General form, Tn = a + (n-1)d
Where Tn is nth term of an Arithmetic Progression
- There are 4 types of questions asked in exams.
Properties of Arithmetic Progression
- If a fixed number is added or subtracted from each term of an AP, then the resulting sequence is also an AP and it has the same common difference as that of the original AP.
- If each term in an AP is divided or multiply with a constant non-zero number, then the resulting sequence is also in an AP.
- If nth term is in linear expression then the sequence is in AP.
- If a1, a2, a3, …, an and b1, b2, b3, …, bn, are in AP. then a1+b1, a2+b2, a3+b3, ……, an+bn and a1–b1, a2–b2, a3–b3, ……, an–bn will also be in AP.
- If nth term of a series is Tn = An + B, then the series is in AP
- Three terms of the A.P whose sum or product is given should be assumed as a-d, a, a+d.
- Four terms of the A.P. whose sum or product is given should be assumed as a-3d, a-d, a+d, a+3d.
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Type 1: Find nth term of series t_{n} = a+(n-1)d
Question 1 : Find 10th term in the series 1, 3, 5, 7, …
Options
A 20
B 19
C 15
D 21
Solution: We know that,
tn = a + (n – 1)d
where tn = nth term,
a= the first term ,
d= common difference,
n = number of terms in the sequence
In the given series,
a (first term) = 1
d (common difference) = 2 (3 – 1, 5 – 3)
Therefore, 10th term = t10 = a + (n-1) d
t10 = 1 + (10 – 1) 2
t10 = 1 + 18
t10 = 19
Correct option: B
Question 2 : Find last term in the series if there are 8 term in this series 13 , 17 , 21 ,25….
Options
A 33
B 41
C 37
D 39
Solution: We know that,
tn = a + (n – 1)d
where tn = nth term,
a = the first term ,
d = common difference,
n = number of terms in the sequence
In the given series,
a (first term) = 13
d (common difference) = 4(17 – 13, 21 – 17)
Therefore, 8th term = t8 = a + (n-1) d
t8 = 13 + (8 – 1) 4
t8 = 13 + 28
t8 = 41
Correct option: B
Type 2: Find number of terms in the series n = \frac{l-a}{d} +1
Question 1 : Find the number of terms in the series 7, 11, 15, . . .71
Options
A 12
B 25
C 22
D 17
Solution: We know that, n= [ \frac{(l-a)}{d} ]+ 1
where n = number of terms,
a= the first term,
l = last term,
d= common difference
In the given series,
a (first term) = 7
l (last term) = 71
d (common difference) = 11 – 7 = 4
n= [\frac{(71-7)}{4}]+ 1
n = \frac{64}{4} + 1
n = 16 + 1
n = 17
Correct option: D
Question 2 : Find the number of terms if First term = 22 ,Last term = 50 and common difference is 4
Options:
A 10
B 9
C 8
D 7
Solution: We know that,
n= [ \frac{(l-a)}{d} ]+ 1where n = number of terms,
a = the first term,
l = last term,
d = common difference
In the given series,
a (first term) = 22
l (last term) = 50
d (common difference) = 4
n = [\frac{(50-22)}{4}]+ 1
n = \frac{28}{4} +1
n = 7+ 1
n = 8
Correct option: C
Type 3: Find sum of first ‘n’ terms of the series S_{n} =\frac{n}{2}[2a+(n-1)\times d] or \frac{n}{2}(a+l)
Question 1. Find the sum of the series 1, 3, 5, 7…. 201
Options
A 12101
B 25201
C 22101
D 10201
Solution: We know that,
Sn = \frac{n}{2}[2a + (n − 1) d]
OR
\frac{n}{2} (a+l)
where,
a = the first term,
d= common difference,
l = tn = nth term = a + (n-1)d
In the given series,
a = 1, d = 2, and l = 201
Since we know that, l = a + (n – 1) d
201 = 1 + (n – 1) 2
201 = 1 + 2n -2
202 = 2n
n = 101
Sn = \frac{ n}{2}(a+l)
Sn = \frac{101}{2} (1+201)
Sn = 50.5 (1 + 201)
Sn = 50.5 x 202
Sn = 10201
Correct option: D
Question 2. Find the sum of the Arithmetic series if First term of this series is 45 , common difference is 5 and number of terms in this series is 8. Options
A 500
B 300
C 400
D 200
Solution: We know that,
Sn = \frac{n}{2}[2a + (n − 1) d]
OR
\frac{n}{2} (a+l)
where, a = the first term,
d = common difference,
l = tn = nth term = a + (n-1)d
In the given series,
a = 45, d = 5, and n = 8
Sn = \frac{n}{2}[2a + (n − 1) d]
Sn = \frac{8}{2}[2\times 45 + (8 − 1) \times 5]
Sn = 4 (90 + 35)
Sn = 4 x 125
Sn = 500
Correct option: A
Type 4: Find the arithmetic mean of the series. b = \frac{a+c}{2}
Question 1.Find the arithmetic mean of first five prime numbers.
Options:
A 6.6
B 3.6
C 5.6
D 7.6
Solution: We know that
b = \frac{1}{2} (a + c)
Here, five prime numbers are 2, 3, 5, 7 and 11
Therefore, their arithmetic mean (AM) = \frac{(2+3+5+7+11)}{5} = 5.6
Correct option: C
Question 2.Find Second Number if arithmetic mean of two numbers is 24 and First number is 10.
Options:
A 32
B 38
C 34
D 36
Solution: We know that
b = \frac{1}{2} (a + c)
Let second Number is b
Therefore, their arithmetic mean (AM) = \frac{(10 + b)}{2} = 24
b = 38
Correct option: B
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plz check your type 3 question 1 answer and option don’t match
Really sorry for the silly mistake Sunishtha, we’ll fix it up