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Progression and Series is one of the most important topics in the CAT Quants section. If you’re weak in Progressions and Series questions for CAT, make sure you learn the basic concepts well. Here, you can learn all the important formulas from CAT Progressions and Series. You can check out these CAT Progression and Series Questions PDF from the CAT Previous year’s papers.Â

This post will look at the important Progressions and Series of questions in the CAT Quant section. These are a good source of practice for CAT preparation; If you want to practice these questions, you can download this CAT Progressions and Series Questions PDF (most important) along with the detailed solutions below, which is completely Free.

Question 1:Â If the product of n positive real numbers is unity, then their sum is necessarily

a)Â a multiple of n

b)Â equal to n + 1/n

c)Â never less than n

d)Â a positive integer

Solution:

Let the numbers be $a_1,a_2….a_n.$

Since the numbers are positive,

$AM\geq GM$

$\frac{a_1+a_2+a_3….+a_n}{n}\geq (a_1*a_2….*a_n)^{1/n}$

$a_1+a_2+a_3….+a_n \geq n$

Question 2:Â If $a_1 = 1$ and $a_{n+1} = 2a_n +5$, n=1,2,….,then $a_{100}$ is equal to:

a)Â $(5*2^{99}-6)$

b)Â $(5*2^{99}+6)$

c)Â $(6*2^{99}+5)$

d)Â $(6*2^{99}-5)$

Solution:

$a_2 = 2*1 + 5$
$a_3 = 2*(2 + 5) + 5 = 2^2 + 5*2 + 5$
$a_4 = 2^3 + 5*2^2 + 5*2 + 5$

$a_{100} = 2^{99} + 5*(2^{98} + 2^{97} + … + 1)$
$= 2^{99} + 5*1*(2^{99} – 1)/(2-1) = 2^{99} + 5*2^{99} – 5 = 6*2^{99} – 5$

Question 3:Â A child was asked to add first few natural numbers (i.e. 1 + 2 + 3 + â€¦) so long his patience permitted. As he stopped, he gave the sum as 575. When the teacher declared the result wrong, the child discovered he had missed one number in the sequence during addition. The number he missed was

a)Â less than 10

b)Â 10

c)Â 15

d)Â more than 15

Solution:

If the child adds all the numbers from 1 to 34, the sum of the numbers would be 1+2+3+…+34 = 34*35/2 = 595
Since the child got the sum as 575, he would have missed the number 20.

Question 4:Â A square is drawn by joining the midpoints of the sides of a given square. A third square is drawn inside the second square in the same way and this process is continued indefinitely. If a side of the first square is 8 cm, the sum of the areas of all the squares such formed (in sq.cm.)is

a)Â 128

b)Â 120

c)Â 96

d)Â None of these

Solution:

Side of first square = 8cm.

Side of second square made by joining mid-points of first square = $\frac{8}{\sqrt2}$

Similarly side of third square =Â Â $\frac{8}{\sqrt2\times\sqrt2}$ and so on.

Now summation of areas will be $8^2+(\frac{8}{\sqrt2})^2+(\frac{8}{\sqrt2\times\sqrt2})^2$ ………

or $8^2 ( 1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}…..)$

or $64 \times (\frac{1}{1-\frac{1}{2}})$ (As we know sum of an infinite G.P. is $\frac{a}{1-r}$ where a is first term and r is common ratio)

or 128 sq. cm.

Question 5:Â If the square of the 7th term of an arithmetic progression with positive common difference equals the product of the 3rd and 17th terms, then the ratio of the first term to the common difference is

a)Â 2:3

b)Â 3:2

c)Â 3:4

d)Â 4:3

Solution:

The seventh term of an AP = a + 6d. Third term will be aÂ + 2d and second term will be aÂ + 16d. We are given that
$(a + 6d)^2 = (a + 2d)(a + 16d)$
=> $a^2$ + $36d^2$ +Â 12ad = $a^2 + 18ad + 32d^2$
=> $4d^2 = 6ad$
=> $d:a = 3:2$

Question 6:Â Let $\ a_{1},a_{2}…a_{52}\$ be positive integers such that $\ a_{1}$ < $a_{2}$ < … <Â $a_{52}\$. Suppose, their arithmetic mean is one less than arithmetic mean of $a_{2}$, $a_{3}$, ….$a_{52}$. If $a_{52}$= 100, then the largest possible value of $a_{1}$is

a)Â 48

b)Â 20

c)Â 23

d)Â 45

Solution:

Let ‘x’ be the average of all 52Â positive integersÂ $\ a_{1},a_{2}…a_{52}\$.

$a_{1}+a_{2}+a_{3}+…+a_{52}$ = 52x … (1)

Therefore, average ofÂ $a_{2}$, $a_{3}$, ….$a_{52}$ = x+1

$a_{2}+a_{3}+a_{4}+…+a_{52}$ = 51(x+1)Â … (2)

From equation (1) and (2), we can say that

$a_{1}+51(x+1)$ =Â 52x

$a_{1}$ = x – 51.

We have to find out the largest possible value ofÂ $a_{1}$.Â $a_{1}$ will be maximum when ‘x’ is maximum.

(x+1) is the average of termsÂ $a_{2}$, $a_{3}$, ….$a_{52}$. We know thatÂ $a_{2}$ < $a_{3}$ < … <Â $a_{52}\$ andÂ $a_{52}$ = 100.

Therefore, (x+1) will be maximum when each term is maximum possible. IfÂ $a_{52}$ = 100, thenÂ $a_{52}$ = 99,Â $a_{50}$ = 98 ends so on.

$a_{2}$ = 100 + (51-1)*(-1) = 50.

Hence, Â $a_{2}+a_{3}+a_{4}+…+a_{52}$ = 50+51+…+99+100 =Â 51(x+1)

$\Rightarrow$ $\dfrac{51*(50+100)}{2} =Â 51(x+1)$

$\Rightarrow$ $x =Â 74$

Therefore,Â theÂ largest possible value ofÂ $a_{1}$ = x – 51 = 74 – 51 = 23.

Question 7:Â Let $a_1, a_2, …$ be integers such that
$a_1 – a_2 + a_3 – a_4 + …. + (-1)^{n – 1} a_n = n,$ for all $n \geq 1.$
Then $a_{51} + a_{52} + …. + a_{1023}$ equals

a)Â 0

b)Â 1

c)Â 10

d)Â -1

Solution:

$a_1 – a_2 + a_3 – a_4 + …. + (-1)^{n – 1} a_n = n$

It is clear from the above equation that when n is odd, the co-efficient of a is positive otherwise negative.

$a_1 – a_2 = 2$

$a_1 = a_2 + 2$

$a_1 – a_2 + a_3 = 3$

On substituting the value of $a_1$ in the above equation, we get

$a_3$ = 1

$a_1 – a_2 + a_3 – a_4 = 4$

On substituting the values of $a_1, a_3$ in the above equation, we get

$a_4$ = -1

$a_1 – a_2 + a_3 – a_4 +a_5 = 5$

On substituting the values of $a_1, a_3, a_4$ in the above equation, we get

$a_5$ = 1

So we can conclude that $a_3, a_5, a_7….a_{n+1}$ = 1 andÂ $a_2, a_4, a_6….a_{2n}$ = -1

Now we have to find the value ofÂ $a_{51} + a_{52} + …. + a_{1023}$

Number of terms = 1023=51+(n-1)1

n=973

There will be 486 even and 487 odd terms, so the value ofÂ $a_{51} + a_{52} + …. + a_{1023}$ = 486*-1+487*1=1

Question 8:Â If $a_1 + a_2 + a_3 + …. + a_n = 3(2^{n + 1} – 2)$, for every $n \geq 1$, then $a_{11}$ equals

Solution:

11th term of series =Â $a_{11}$ = Sum of 11 terms – Sum of 10 terms = $3(2^{11 + 1} – 2)$-3$(2^{10 + 1} – 2)$

= 3$(2^{12} – 2-2^{11} +2)$=3$(2^{11})(2-1)$= 3*$2^{11}$ = 6144

Question 9:Â Three positive integers x, y and z are in arithmetic progression. If $y-x>2$ and $xyz=5(x+y+z)$, then z-x equals

a)Â 8

b)Â 12

c)Â 14

d)Â 10

Solution:

Given x, y, z are three terms in an arithmetic progression.

Considering x = a, y = a+d, z = a+2*d.

Using the given equation x*y*z = 5*(x+y+z)

a*(a+d)*(a+2*d) = 5*(a+a+d+a+2*d)

=a*(a+d)*(a+2*d)Â  = 5*(3*a+3*d) = 15*(a+d).

= a*(a+2*d) = 15.

Since all x, y, z are positive integers and y-x > 2. a, a+d, a+2*d are integers.

The common difference is positive and greater than 2.

Among the different possibilities are : (a=1, a+2d = 5), (a, =3, a+2d = 5), (a = 5, a+2d = 3), (a=15, a+2d = 1)

Hence the only possible case satisfying the condition is :

a = 1, a+2*d = 15.

x = 1, z = 15.

z-x = 14.

Question 10:Â Consider the arithmetic progression 3, 7, 11, … and let $A_n$ denote the sum of the first n terms of this progression.Â Then the value of $\frac{1}{25} \sum_{n=1}^{25} A_{n}$ is

a)Â 455

b)Â 442

c)Â 415

d)Â 404

Solution:

Sum of n terms in an A.P =Â $\frac{n}{2}\left(2a+\left(n-1\right)d\right)$

$A_n=\frac{n}{2}\left(6+\left(n-1\right)4\right)=n\left(2n+1\right)$

$\Sigma\ A_n=\Sigma\ n\left(2n+1\right)=2\Sigma\ n^2+\Sigma\ n=\ \frac{\ 2n\left(n+1\right)\left(2n+1\right)}{6}+\ \frac{\ n\left(n+1\right)}{2}$

Substituting n = 25, we get

$\frac{1}{25} \sum_{n=1}^{25} A_{n}$ =Â $\frac{1}{25}\left(\ \frac{\ 2\left(25\right)\left(25+1\right)\left(50+1\right)}{6}+\ \frac{\ 25\left(25+1\right)}{2}\right)$

$\frac{1}{25} \sum_{n=1}^{25} A_{n}$ =Â $\ 26\left(17\right)+13$ = 455