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# Series Questions for NMAT:

Download Series Questions for NMAT PDF. Top 10 very important Series Questions for NMAT based on asked questions in previous exam papers.

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Question 1:Â If $a_1, a_2, ……$ are in A.P., then, $\frac{1}{\sqrt{a_1} + \sqrt{a_2}} + \frac{1}{\sqrt{a_2} + \sqrt{a_3}} + ……. + \frac{1}{\sqrt{a_n} + \sqrt{a_{n + 1}}}$ is equal to

a)Â $\frac{n}{\sqrt{a_1} + \sqrt{a_{n + 1}}}$

b)Â $\frac{n – 1}{\sqrt{a_1} + \sqrt{a_{n – 1}}}$

c)Â $\frac{n – 1}{\sqrt{a_1} + \sqrt{a_n}}$

d)Â $\frac{n}{\sqrt{a_1} – \sqrt{a_{n + 1}}}$

Question 2:Â If p, q and r are three unequal numbers such that p, q and r are in A.P., and p, r-q and q-p are in G.P., then p : q : r is equal to:

a)Â 1 : 2 : 3

b)Â 2 : 3 : 4

c)Â 3 : 2 : 1

d)Â 1 : 3 : 4

Question 3:Â If the positive real numbers a, b and c are in Arithmetic Progression, such that abc = 4, then minimum possible value of b is:

a)Â $2^{\frac{3}{2}}$

b)Â $2^{\frac{2}{3}}$

c)Â $2^{\frac{1}{3}}$

d)Â None of the above

Question 4:Â The interior angles of a polygon are in Arithmetic Progression. If the smallest angle is 120Â° and common difference is 5Â°, then number of sides in the polygon is:

a)Â 7

b)Â 8

c)Â 9

d)Â None of the above

Question 5:Â Sum of the series $1^{2} – 2^{2} + 3^{2} – 4^{2} + … + 2001^{2} – 2002^{2} + 2003^{2}$ is:

a)Â 2007006

b)Â 1005004

c)Â 200506

d)Â None of the above

Question 6:Â The sum of 4+44+444+…. upto n terms in

a)Â $\frac{40}{81}(8^{n}-1)-\frac{5n}{9}$

b)Â $\frac{40}{81}(8^{n}-1)-\frac{4n}{9}$

c)Â $\frac{40}{81}(10^{n}-1)-\frac{4n}{9}$

d)Â $\frac{40}{81}(10^{n}-1)-\frac{5n}{9}$

Question 7:Â The sum of series, (-100) + (-95) + (-90) + â€¦â€¦â€¦â€¦+ 110 + 115 + 120, is:

a)Â 0

b)Â 220

c)Â 340

d)Â 450

e)Â None of the above

Question 8:Â The sum of 3rd and 15th elements of an arithmetic progression is equal to the sum of 6th, 11th and 13th elements of the same progression. Then which element of the series should necessarily be equal to zero?

a)Â 1st

b)Â 9th

c)Â 12th

d)Â None of the above

Question 9:Â The 288th term of the series a,b,b,c,c,c,d,d,d,d,e,e,e,e,e,f,f,f,f,f,fâ€¦ is

a)Â u

b)Â v

c)Â w

d)Â x

Question 10:Â The nth element of a series is represented as
$X_n = (-1)^nX_{n-1}$
If $X_0 = x$ and $x > 0$, then which of the following is always true?

a)Â $X_n$ is positive if n is even

b)Â $X_n$ is positive if n is odd

c)Â $X_n$ is negative if n is even

d)Â None of these

We have,Â $\frac{1}{\sqrt{a_1} + \sqrt{a_2}} + \frac{1}{\sqrt{a_2} + \sqrt{a_3}} + ……. + \frac{1}{\sqrt{a_n} + \sqrt{a_{n + 1}}}$

Now,Â $\frac{1}{\sqrt{a_1} + \sqrt{a_2}}$Â =Â $\frac{\sqrt{a_2} – \sqrt{a_1}}{(\sqrt{a_2} + \sqrt{a_1})(\sqrt{a_2} – \sqrt{a_1})}$Â  Â (Multiplying numerator and denominator byÂ $\sqrt{a_2} – \sqrt{a_1}$)

= $\frac{\sqrt{a_2} – \sqrt{a_1}}{({a_2} – {a_1}}$

=$\frac{\sqrt{a_2} – \sqrt{a_1}}{d}$Â  Â (where d is the common difference)

Similarly,Â $\frac{1}{\sqrt{a_2} + \sqrt{a_3}}$ =Â $\frac{\sqrt{a_3} – \sqrt{a_2}}{d}$ and so on.

Then the expressionÂ $\frac{1}{\sqrt{a_1} + \sqrt{a_2}} + \frac{1}{\sqrt{a_2} + \sqrt{a_3}} + ……. + \frac{1}{\sqrt{a_n} + \sqrt{a_{n + 1}}}$

can be written asÂ $\ \frac{\ 1}{d}(\sqrt{a_2}-\sqrt{a_1}+\sqrt{a_3}-\sqrt{a_3}+……………………..\sqrt{a_{n+1}} – \sqrt{a_{n}}$

=Â $\ \frac{\ n}{nd}(\sqrt{a_{n+1}}-\sqrt{a_1})$Â (Multiplying both numerator and denominator by n)

= $\ \frac{n(\sqrt{a_{n+1}}-\sqrt{a_1})}{{a_{n+1}} – {a_1}}$Â  Â  Â $(a_{n+1} – {a_1} =nd)$

=Â $\frac{n}{\sqrt{a_1} + \sqrt{a_{n + 1}}}$

Given that p, q and r are in A.P.,

2q = p + r

p = 2q – rÂ  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â Eq -1

Given that p, r-q and q-p are in G.P.,

Let us assume the common ratio of k in G.P.

r-q = k(p)Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â Eq -2

q-p = k(r-q)Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â Eq -3

q-p = $k^{2}$(p)Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â Eq -4

Substitute Eq-1 in Eq-3,

q-(2q-r) = k(r-q)

r-q = k(r-q)

So, k=1

From Eq -4, we get q=2p

Now substitute q=2p in Eq-1 we get r=3p

Hence, ratio of p:q:r = p:2p:3p = 1:2:3

It has been given that a, b, and c are in an arithmetic progression.
Let a = x-p, b = x, and c = x+p
We know that a, b, and c are real numbers.
Therefore, the arithmetic mean of a,b,c should be greater than or equal to the geometric mean.
$\frac{a+b+c}{3} \geq \sqrt[3]{abc}$
$\frac{a+b+c}{3} \geq \sqrt[3]{4}$
$\frac{3x}{3}\geq\sqrt[3]{4}$

$x\geq\sqrt[3]{4}$
We know that $x$ = $b$.
Therefore,$b\geq\sqrt[3]{4}$orÂ $b\geq 2^{\frac{2}{3}}$
Therefore, optionÂ B is the right answer.

It has been given that the interior angles in a polygon are in an arithmetic progression.
We know that the sum of all exterior angles of a polygon is 360Â°.
Exterior angle = 180Â° – interior angle.
Since we are subtracting the interior angles from a constant, the exterior angles will also be in an AP.
The starting term of the AP formed by the exterior angles will be 180Â°-120Â° = 60Â° and the common difference will be -5Â°.

Let the number of sides in the polygon be ‘n’.
=> The number of terms in the series will also be ‘n’.

We know that the sum of an AP is equal to 0.5*n*(2a + (n-1)d), where ‘a’ is the starting term and ‘d’ is the common difference.
0.5*n*(2*60Â° + (n-1)*(-5Â°)) = 360Â°
120$n$ – 5$n^2$ + 5$n$ = 720
5$n^2$ – 125$n$ + 720 = 0
$n^2$ – 25$n$ + 144 =0.
$(n-9)(n-16) = 0$

Therefore, $n$ can be 9 or 16.
If the number of sides is 16, then the largest external angle will be 60 – 15*5 = -15Â°. Therefore, we can eliminate this case.
The number of sides in the polygon must be 9. Therefore, option C is the right answer.

The given series is $1^2 – 2^2 + 3^2 – 4^2 +…..+2003^2$
$1^2 – 2^2$ can be written as $(1+2)(1-2)$ = $3*(-1)$ = $-3$
$3^2 -4^2$ can be written as $(3+4)*(3-4)$ = $7*(-1)$ = $-7$
$5^2-6^2$ can be written as $(5+6)*(5-6)$ = $11*(-1)$ = $-11$
Therefore, all the terms till $2002^2$ can be expressed as an AP.
The last term of the AP will be $(2001+2002)(2001-2002)$ = $-4003$

Therefore, the given expression is reduced to $-3 – 7 …-4003 + 2003^2$
Let is evaluate the value ofÂ $-3 – 7 …-4003$
Number of terms,$n$ = $\frac{4003-3}{4} + 1$ = $1001$
Sum = $\frac{n}{2} *$(first term + last term)
= $\frac{1001}{2}*(-4006)$
= $-2005003$
$2003^2 = 4012009$
Value of the given expression = $4012009 – 2005003 = 2007006$.

Therefore, option A is the right answer.

Given that, S = 4+44+444+…

$\Rightarrow$ $S = \frac{4}{9}(9+99+999+…)$

$\Rightarrow$Â $S = \frac{4}{9}(10-1+10^2-1+10^3-1+…+10^n-1)$

$\Rightarrow$Â $S = \frac{4}{9}(10+10^2+10^3+…+10^n-n)$

$\Rightarrow$Â $S = \frac{4}{9}(\frac{10(10^n – 1)}{10-1}-n)$

$\Rightarrow$Â $S = \frac{40}{81}(10^n – 1) – \frac{4n}{9}$

Hence, option C is the correct answer.

Alternate method:

Solving for n = 2

Sum of series = 4+44 = 48

Substituting n = 2 in options

(A)Â $\frac{40}{81}(8^{2}-1)-\frac{5*2}{9}$ = $\frac{280-10}{9}$ = 30

(B)Â $\frac{40}{81}(8^{2}-1)-\frac{4*2}{9}$ = $\frac{280-8}{9}$ =Â Â $\frac{272}{9}$

(C) $\frac{40}{81}(10^{2}-1)-\frac{4*2}{9}$ =Â $\frac{440-8}{9}$ = 48

(D)Â $\frac{40}{81}(10^{2}-1)-\frac{5*2}{9}$ =Â $\frac{440-10}{9}$ = $\frac{430}{9}$

The given series is (-100)+ (-95)+ (-90)+….+110 +Â 115 + 120.
We can observe that -100 will cancel out 100, -95 will cancel out 95 and so on. Therefore, the only terms that will be remaining are 105, 110, 115 and 120.
Sum of the series = 105 + 110 +Â 115 +Â 120 = 450.
Therefore, option D is the right answer.

The sum of the 3rd and 15th terms is a+2d+a+14d = 2a+16d
The sum of the 6th, 11th and 13th terms is a+5d+a+10d+a+12d = 3a+27d
Since the two are equal, 2a+16d = 3a+27d => a+11d = 0
So, the 12th term is 0

1, 2, 3, 4,….n such that the sum is greater than 288
If n = 24, n(n+1)/2 = 12*25 = 300
So, n = 24, i.e. the 24th letter in the alphabet is the letter at position 288 in the series
Let x = 1, so, $X_0$ = 1
$X_1$ = -1
$X_2$ = -1
$X_3$ = 1
$X_4$ = 1
$X_5$ = -1
$X_6$ = -1
So,Â $X_n$ need not be positive when n is even,Â $X_n$ need not be positive when n is odd,Â $X_n$ need not be negative when n is even. So, none of the first three options are correct.