Progression – Aptitude Questions and Answers

Last Updated : 08 Nov, 2024

Progression (or Sequences and Series) are mathematical concepts that involve arranging numbers in a particular order based on a repeatable pattern. The topic of Progressions is frequently asked in various competitive exams like SSC, Bank PO, and other government job exams and is a crucial part of Quantitative Aptitude, but it can be mastered with the right formulas and working through some examples.

Prerequisites:

Aptitude Questions on Progression

Q1: Find the nth term for the AP: 11, 17, 23, 29, …

Solution:

Here, a = 11, d = 17 – 11 = 23 – 17 = 29 – 23 = 6We know that nth term of an AP is a + (n – 1) d=> nth term for the given AP = 11 + (n – 1) 6=> nth term for the given AP = 5 + 6 nWe can verify the answer by putting values of ‘n’.=> n = 1 -> First term = 5 + 6 = 11=> n = 2 -> Second term = 5 + 12 = 17=> n = 3 -> Third term = 5 + 18 = 23and so on …

Q2: Find the sum of the AP in the above question till the first 10 terms.

Solution :

From the above question, => nth term for the given AP = 5 + 6 n=> First term = 5 + 6 = 11=> Tenth term = 5 + 60 = 65=> Sum of 10 terms of the AP = 0.5 n (first term + last term) = 0.5 x 10 (11 + 65)=> Sum of 10 terms of the AP = 5 x 76 = 380

Q3: For elements 4 and 6, verify that A ≥ G ≥ H.

Solution :

A = Arithmetic Mean = (4 + 6) / 2 = 5G = Geometric Mean = [Tex]\sqrt{{4}\times{6}} [/Tex]= 4.8989H = Harmonic Mean = (2 x 4 x 6) / (4 + 6) = 48 / 10 = 4.8Therefore, A ≥ G ≥ H

Q4: Find the sum of the series 32, 16, 8, 4, … upto infinity.

Solution :

First term, a = 32Common ratio, r = 16 / 32 = 8 / 16 = 4 / 8 = 1 / 2 = 0.5We know that for an infinite GP, Sum of terms = a / (1 – r)=> Sum of terms of the GP = 32 / (1 – 0.5) = 32 / 0.5 = 64

Q5: The sum of three numbers in a GP is 26 and their product is 216. ind the numbers.

Solution:

Let the numbers be a/r, a, ar.=> (a / r) + a + a r = 26=> a (1 + r + r²) / r = 26Also, it is given that product = 216=> (a / r) x (a) x (a r) = 216=> a³ = 216=> a = 6=> 6 (1 + r + r²) / r = 26=> (1 + r + r²) / r = 26 / 6 = 13 / 3=> 3 + 3 r + 3 r² = 13 r=> 3 r² – 10 r + 3 = 0=> (r – 3) (r – (1 / 3) ) = 0=> r = 3 or r = 1 / 3Thus, the required numbers are 2, 6 and 18.

Q6: Find the middle term of the A.P. 6, 13, 20, … , 216.

Solution:

In the given AP,

First term a₁ = 6
Last term a_n = 216
Common difference = 7
Now, to find the number of terms, n = (a_n – a₁)/d + 1

n = (216 – 6)/7 + 1

n = 210/7 + 1

n = 30 + 1

n = 31

So, middle term is (n + 1) / 2

= (31 + 1)/2

= 16

Now to calculate the middle term

a₁₆ = a1 + 15×d

a₁₆ = 6 + 15 × 7

a₁₆ = 6 + 105 = 111

So the middle term of the given AP is 111

Q7: In an AP, if ? = −2, ? = 5 and ?? = 0, then find the value of ? ?

Solution:

Given values are:

d = -2
n = 5
a_n = 0
So, to calculate follow these steps:

a_n = a + (n-1)d

0 = a + (5-1)(-2)

0 = a – 8

a = 8

Q8: Number of bacteria in a dish are 100, and they are increasing by double the previous value every hour. Find the number of bacteria in the dish after 6 hours.

Solution:

Here, every year the number becomes 2 times. A constant number is being multiplied to the previous term to get the new term. This is a geometric progression.

100,200, 400 … and so on.

Here a = 100 and r= 2

Using the formula for sum till nth term of the GP

S_n = a(1–rn)1–r1–ra(1–rn)

n = 6. Plugging in the values in the formula

S_n = a(1–rn)1–r1–ra(1–rn)

⇒S_n = a(1–rn)1–r1–ra(1–rn)

⇒ S₆ = 100(1–26)1–21–2100(1–26)

⇒ S₆ = 100(26–1)2–12–1100(26–1)

⇒ S₆ = 100(64–1)2–12–1100(64–1)

⇒ S₆ = 6300

There must be 63,00 bacteria in the dish now.

Also Read,

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Profit and Loss Aptitude Questions

Practice Questions on Progression

Question 1. Find the next number in the sequence: 2, 5, 10, 17, 26, …

Question 2. In an arithmetic progression, the 5th term is 18 and the 9th term is 30. Find the first term and the common difference.

Question 3. The sum of the first n terms of an AP is 3n² + 5n. Find the nth term of this AP.

Question 4. Find the 8th term of the GP: 3, 6, 12, 24, …

Question 5. The sum of the first three terms of a GP is 26, and the sum of their squares is 364. Find the terms.

Question 6. In an AP, the sum of the first 5 terms is 75 and the sum of the next 5 terms is 125. Find the first term and the common difference.

Question 7. If the 4th and 7th terms of a GP are 54 and 432 respectively, find the common ratio and the first term.

Question 8. Find the sum of all three-digit numbers which are divisible by 7.

Question 9. The sum of n terms of the series 5 + 55 + 555 + … is 6170. Find n.

Question 10. In a sequence, each term after the first is found by adding 1 to the product of the two preceding terms. If the first two terms are 2 and 3, what is the 5th term?