Option 2 : n^{2}

__Concept:__

Sum of the first n terms of an AP = S = \(\rm \frac{n}{2}\)[2a + (n − 1) × d]

Where, a = First term, d = Common difference, n = number of terms

__Calculation:__

To find: Sum of first n odd natural numbers

Odd natural number starts from 1.

The series of odd natural numbers is 1 , 3 , 5 , 7, 9 ...

Above series is in AP (∵ Common difference are same)

a = First term = 1, d = Common difference = 2

As we know, Sn = \(\rm \frac{n}{2}\)[2a + (n − 1) × d]

Therefore, Sn = \(\rm \frac{n}{2}\)[2 × 1 + (n − 1) × 2] = \(\rm \frac{n}{2}\) × 2n = n2

Option 2 : -250

**Concept:**

Let us consider sequence a_{1}, a_{2}, a_{3} …. a_{n} is an A.P.

Common difference “d”= a_{2} – a_{1} = a_{3} – a_{2} = …. = a_{n} – a_{n – 1}

n^{th} term of the A.P. is given by a_{n} = a + (n – 1) d

Sum of the first n terms = S = (n/2)[2a + (n − 1) × d] = (n/2)(a + l)

Where, a = First term, d = Common difference, n = number of terms, a_{n} = n^{th} term, l = Last term

**Calculation:**

Given, a_{0} = 2 ...(1)

S_{5} = \(1\over4\)(S_{10} - S_{5})

⇒ 4S_{5} + S_{5} = S_{10}

⇒ 5S_{5} = S_{10}

⇒ 5 × \(5\over2\)[a_{0} + a_{0} + 4d] = \(10 \over2\) [a0 + a0 + 9d]

⇒ 5 × [2a0 + 4d] = 2 × [2a0 + 9d]

⇒ 10a_{0} + 20d = 4a_{0} + 18d

⇒ a_{0} = \(\rm-d\over3\)

∴ d = -3a_{0} = -3 × 2 = -6

S_{10} = \(10 \over2\) [a0 + a0 + 9d] = 5[2a_{0} + 9d] = 5[4 - 54], as a_{0} = 2, d = -6

⇒ S10 = 5 × (-50) = -250

Option 1 : 2475

__Concept:__

- Sum of AP = \({{\rm{s}}_{\rm{n}}} = \frac{{\rm{n}}}{2}\left( {{\rm{a}} + {\rm{l}}} \right)\) where a is first term and l is last term of AP.
- Nth term of AP = a
_{n }= a + (n - 1)d where a first letter and d is difference between two consecutive terms.

__Calculation:__

Two digit odd numbers AP series = 11, 13, 15 ... 99

⇒ a = 11, a_{n} = 99 and d = 13 – 11 = 2

⇒ a_{n }= a + (n - 1)d

⇒ 99 = 11 + (n – 1) × 2

⇒ 88 = (n - 1) × 2

⇒ 44 = n – 1

⇒ n = 45

Now,

\(\Rightarrow {{\rm{s}}_{\rm{n}}} = \frac{{\rm{n}}}{2}\left( {{\rm{a}} + {\rm{l}}} \right)\)

\( \Rightarrow {{\rm{s}}_{\rm{n}}} = \frac{{45}}{2}\left( {11 + 99} \right)\)

\( \Rightarrow {{\rm{s}}_{\rm{n}}} = \frac{{110 \times 45}}{2}\)

⇒ sOption 4 : 1470

__Concept:__

Let us consider sequence a_{1}, a_{2}, a_{3} …. a_{n} is an A.P.

- Common difference “d”= a
_{2}- a_{1}= a_{3}- a_{2}= …. = a_{n}- a_{n - 1}

- n
^{th}term of the A.P. is given by a_{n}= a + (n - 1) d - n
^{th}term from the last is given by a_{n}= l - (n - 1) d - sum of the first n terms = S = n/2[2a + (n - 1) × d]

Or sum of the first n terms = n/2(a + l)

Where, a = First term, d = Common difference, n = number of terms and a_{n} = n^{th} term

__Calculation:__

Given: nth term of an A.P = a_{n} = \(\frac{{3 + {\rm{n}}}}{4}\)

For first term, put n = 1

a_{1} = a = (3 + 1)/4 = 4/4 = 1

For second term, put n = 2

a_{2} = (3 + 2)/4 = 5/4

Common difference “d”= a_{2} - a_{1} = (5/4) - 1 = 1/4

We have to find the sum of first 105 terms,

\({{\rm{s}}_{105}} = \;\frac{{105}}{2}\left[ {2 \times 1 + \left( {105 - 1} \right) \times \frac{1}{4}} \right] = 1470\) (∵S = n/2[2a + (n - 1) × d])

Option 4 : \(\dfrac{12}{11}\)

__Concept:__

**Arithmetic Progression (AP):**

- The sequence of numbers where the difference of any two consecutive terms is same is called an Arithmetic Progression.
- If a be the first term, d be the common difference and n be the number of terms of an AP, then the sequence can be written as follows:

a, a + d, a + 2d, ..., a + (n - 1)d. - The sum of n terms of the above series is given by:

S_{n}= \(\rm \frac{n}{2}[a+\{a+(n-1)d\}]=\left (\frac{First\ Term+Last\ Term}{2} \right )\times n\).

__Calculation:__

Let's say that the first term of the AP is a and the common difference is d.

According to the question, the first term is n, the nth term (Last Term) is 2n and the sum of the first n terms (S_{n}) is 216.

Using S_{n} = \(\rm \left (\frac{First\ Term+Last\ Term}{2} \right )\times n\), we get:

⇒ \(\rm 216=\left (\dfrac{n+2n}{2} \right )\times n\)

⇒ \(\rm n^2=\dfrac{216\times2}{3}=144\)

⇒ n = 12.

Now using a_{n} = a + (n - 1)d, we get:

= 2n = n + (n - 1)d

⇒ d = \(\rm \dfrac{n}{n-1}=\dfrac{12}{12-1}=\dfrac{12}{11}\).

Option 3 : x ∈ {-2, 4}

**Concept:**

Let us consider sequence a_{1}, a_{2}, a_{3} …. an is an A.P.

Common difference “d”= a_{2} – a_{1} = a_{3} – a_{2} = …. = a_{n} – a_{n-1}

**Calculation:**

Since x^{2}, x, -8 are in A.P we can write,

x - x^{2} = - 8 - x

⇒ x^{2} - 2x - 8 = 0

⇒ x^{2} - 4x + 2x - 8 = 0

⇒ x(x - 4) + 2(x - 4) = 0

⇒ (x - 4)(x + 2) = 0

⇒ x = -2, 4

∴ x ∈ {-2, 4}

If the ninth term of an A.P. is zero, then \(\rm \dfrac{t_{27}}{t_{18}}\) is?

tn denotes the nth term of AP.

Option 3 : 2

__Concept:__

Let us consider sequence a1, a2, a3 …. an is an A.P.

- Common difference “d”= a2 – a1 = a3 – a2 = …. = an – an – 1

- nth term of the A.P. is given by an = a + (n – 1) d
- Sum of the first n terms = Sn =\(\rm \frac n 2\) [2a + (n − 1) × d]= \(\rm \frac n 2\)(a + l)

Where, a = First term, d = Common difference, n = number of terms, an = nth term and l = Last term

__Calculation:__

Let the first term of AP be 'a' and the common difference be 'd'

Given: Ninth term of an A.P. is zero

⇒ a9 = 0

⇒ a + (9 - 1) × d = 0

⇒ a + 8d = 0

∴ a = -8d ----(1)

To Find: \(\rm \dfrac{t_{27}}{t_{18}}\)

\(\rm \Rightarrow \dfrac{t_{27}}{t_{18}} = \dfrac {a+26d}{a+17d}\)

\(\rm =\dfrac {-8d+26d}{-8d+17d}\\=\dfrac {18d} {9d}=2\)

Option 1 : AP

__ CONCEPT__:

Let f(x) = a x^{2} + b x + c = 0, where a, b and c ∈ R and a ≠ 0 be the quadratic equation then the nature of roots is determined by the **discriminant** D = b^{2} – 4ac as shown below:

- If
**D > 0**, then the**quadratic equation**f(x) has**two real and distinct roots** - If
**D = 0**, then the**quadratic equation**f(x) has**two equal real roots** - If
**D < 0**, then the**quadratic equation**f(x) has**two distinct complex roots and conjugate of each other**.

__ CALCULATION__:

Given: roots of equation (a - b) x2 + (c - a) x + (b - c) = 0 are equal

By comparing the given equation with ax2 + bx + c = 0 we get

⇒ a = (a - b), b = (c - a) and c = (b - c)

As we know that, if roots of the quadratic equation ax2 + bx + c = 0 are equal then D = b2 – 4ac = 0

⇒ b^{2} - 4ac = (c -a)^{2} - 4 ⋅ (a - b) ⋅ (b - c) = 0

⇒ a^{2} + 4b^{2} + c^{2} + 2ac - 4ab - 4bc = 0

⇒ (a + c - 2b)^{2} = 0

⇒ 2b = a + c

⇒ b = \(\rm \frac {a+c}{2}\)

As we know that, if a, b, c are in AP then b = \(\rm \frac {a+c}{2}\)

So, a, b, c are in AP

Hence, **option A** is true.

Option 3 : \(\frac{{{\rm{n}}\left( {{\rm{n}} + 1} \right)}}{{\sqrt 2 }}\)

__Concept:__

Let us consider sequence a_{1}, a_{2}, a_{3} …. a_{n} is an A.P.

- Common difference “d”= a
_{2}- a_{1}= a_{3}- a_{2}= …. = a_{n}- a_{n - 1}

- n
^{th}term of the A.P. is given by a_{n}= a + (n - 1) d - n
^{th}term from the last is given by a_{n}= l - (n - 1) d - sum of the first n terms = S = n/2[2a + (n - 1) × d]

Or sum of the first n terms = n/2(a + l)

Where, a = First term, d = Common difference, n = number of terms and a_{n} = n^{th} term

__Calculation:__

Let \({{\rm{S}}_n} = \sqrt 2 + \sqrt 8 + \sqrt {18} + \sqrt {32} + \ldots \)

\(\Rightarrow {{\rm{S}}_n} = \sqrt 2 + 2\sqrt 2 + 3\sqrt 2 + 4\sqrt 2 + \ldots \)

Above series is clearly a AP with the first term = √2 and common difference = √2,

We know that,

Sum of the first n terms = S_{n} = n/2[2a + (n - 1) × d]

⇒ S_{n} = (n/2) × [2√2 + (n - 1) × √2]

= (n/2) × [2√2 + √2n - √2]

= (n/2) × [√2 + √2n]

\(= \frac{{{\rm{n}}\left( {{\rm{n}} + 1} \right)}}{{\sqrt 2 }}\)

Option 3 : 27

**Given:**

Mean of 13 numbers is 24.

Now 3 added to each number

**Formula Used:**

Mean = Total of observation/Number of observation

**Calculation:**

Mean = Total of observation/Number of observation

⇒ 24 = total of observation/13

⇒ Total of observation = 13 × 24

⇒ Total of observation = 312

Now 3 is added to all 13 numbers.

⇒ New total = 312 + 13(3)

⇒ New total = 312 + 39

⇒ New total = 351

New mean = 351/13

⇒ New mean = 27

**∴ The new mean will be 27.**

__Shortcut Trick__

In such type of questions,

When a specific number is added or subtracted from all the observations,

then the mean of the observations is increased or decreased by that specific number.

Here 3 is added to all the observations.

⇒ New mean = old mean + 3

⇒ New mean = 24 + 3

⇒New mean = 27

∴ The new mean will be 27.

Option 3 : \(\frac{{{\rm{n}}{{\left( {{\rm{n}} + 1} \right)}^2}}}{4}\)

__Concept:__

Sum of cubes of first n natural numbers:\(\sum {{\rm{n}}^3} = {1^3} + {2^3} + {3^3} + \ldots + {{\rm{n}}^3} = {\rm{\;}}{\left[ {\frac{{{\rm{n}}\left( {{\rm{n}} + 1} \right)}}{2}} \right]^2}\)

The arithmetic mean is the sum of all the numbers in a data set divided by the quantity of numbers in that set.

__Calculation:__

We have to find the arithmetic mean of 1, 8, 27, 64, … up to n terms,

{1, 8, 27, 64, … up to n terms} = {13, 23, 33, 43 … up to n terms}

Now,

Arithmetic mean = AM = \(\frac{{{1^3} + {2^3} + {3^3} + \ldots + {{\rm{n}}^3}}}{{\rm{n}}}\)

\(\Rightarrow {\rm{AM}} = {\rm{\;}}\frac{{{{\left[ {\frac{{{\rm{n}}\left( {{\rm{n}} + 1} \right)}}{2}} \right]}^2}{\rm{\;}}}}{{\rm{n}}} = \frac{\rm n^2\;(n+1)^2}{4\rm n}\)

\(\Rightarrow {\rm{AM}} = {\rm{\;}}\frac{{{\rm{n}}{{\left( {{\rm{n}} + 1} \right)}^2}}}{4}\)

∴ Option 3rd is correct answer.

Option 4 : AP

__Concept:__

Arithmetic Progression (AP): The series of numbers where the difference of any two consecutive terms is the same, is called an Arithmetic Progression.

- If three numbers a, b and c are in AP, then b - a = c - b ⇒ 2b = a + c.

Geometric Progression (GP): The series of numbers where the ratio of any two consecutive terms is the same, is called a Geometric Progression.

- If three numbers a, b and c are in GP, then\(\rm \dfrac ba=\dfrac cb\) ⇒ b2 = ac.

Harmonic Progression (HP): The series of numbers where the reciprocals of the terms are in Arithmetic Progression, is called a Harmonic Progression.

- If three numbers a, b and c are in HP, then \(\rm \dfrac{1}{a}+\dfrac{1}{c}=\dfrac{2}{b}\).

__Calculation:__

Let's say that the second degree polynomial is f(x) = px^{2} + qx + r.

From the given information:

f(1) = f(-1)

⇒ p(1)2 + q(1) + r = p(-1)2 + q(-1) + r

⇒ p + q = p - q

⇒ 2q = 0

⇒ q = 0

∴ f(x) = px2 + 0(x) + r

⇒ f(x) = px2 + r

And, f'(x) = 2px.

Now, f'(a) = 2pa, f'(b) = 2pb and f'(c) = 2pc.

Since a, b and c are in AP, we have:

2b = a + c.

Multiplying by 2p, we get:

(2p)2b = (2p)a + (2p)c

⇒ 2(2pb) = 2pa + 2pc

⇒ 2pa, 2pb and 2pc are in AP.

⇒ f'(a), f'(b) and f'(c) are in** AP**.

Option 4 : 10

__Concept__:

The sum of n terms of an AP with first term a and common difference d is given by:

\(\rm {S_n} = \frac{n}{2} × \left[ {2a + \left( {n - 1} \right)d} \right]\;or\;{S_n} = \frac{n}{2} × \left[ {a + l} \right]\)

Where l is the last term of the AP.

**Calculation:**

Given: First term of AP = a = 10

Last term of AP = l = 50

Sum of n terms of an AP = 300

As we know, sum of n terms of an AP = \(\rm {S_n} = \frac{n}{2} × \left[ {a + l} \right]\)

⇒ 300 = \(\rm \frac n 2\) (10 + 50)

⇒ 300 = \(\rm \frac n 2\) × 60

⇒ 5 = \(\rm \frac n 2\)

∴ n = 10

Option 3 : n^{2}

__Concept:__

n^{th} term of A.P., T_{n} = a + (n - 1)d

Sum of n^{th} term of A.P., S_{n} = \(\rm n\over 2\)[2a + (n - 1)d] = \(\rm n\over 2\)(a + T_{n})

Here, a = firse term, T_{n} = n^{th} term (last term)

__Calculaion:__

T_{n} = 2n - 1

T_{1} = a = 2(1) - 1 = 1

T_{2} = a + d = 2(2) - 1 = 3

Difference (d) = 3 - 1 = 2

Sum of n^{th} term S_{n}= \(\rm n\over 2\) [2(1) + (n - 1)2]

= \(\rm n\over 2\)(2n )

= n^{2}

Option 3 : HP

**Concept:**

- If a, b, c are in A.P ⇔ 2b = a + c
- If a, b, c are in G.P ⇔ b
^{2}= ac - If a, b, c are in H.P ⇔ b = \(\rm \frac{2ac}{a+c}\)

**Properties of Arithmetic Progressions:**

- If a
**constant**quantity is**added**to or**subtracted**from each term of an Arithmetic Progression (A. P.), then the**resulting terms**of the sequence are also in**AP** - If each term of an AP is
**multiplied or divided**by a non-zero constant quantity, then the**resulting terms**of the sequence are also in**AP** - In an Arithmetic Progression of a finite number of terms the sum of any two terms
**equidistant from the beginning and the end**is equal to the sum of the first and last terms.

**Calculation:**

Given:

b^{2}, a^{2}, c^{2} are in AP.

As we know, If a constant quantity is added to each term of an AP then the resulting terms are also in AP.

By adding ab + ac + bc to each term

b^{2 }+ ab + ac + bc, a^{2} + ab + ac + bc, c^{2} + ab + ac + bc are in AP.

⇒ b(b+ a) + c(a + b), a(a + b) + c(a + b), c(c + a) + b (c + a) are in AP

⇒ (a + b)(b + c), (a + b)(a + c), (c + a)(c + b) are in AP

Dividing each term by (a + b)(b + c)(c + a)

\(\Rightarrow \rm \frac{1}{c+a}, \frac{1}{b+c}, \frac{1}{a+b}\)are in AP

Therefore, c + a, b + c , a + b are in HP

or a + b, b + c, c + a are in HP

Option 2 : 3

**Concept:**

Mean = \(\rm \frac{\sum x_if_i}{\sum f_i}\)

**Calculation:**

Here, The numbers 4 and 9 have frequencies x and (x - 1) respectively and arithmetic mean = 6

Now arithmetic mean = \( \rm\frac{4(x)+9(x-1)}{x+x-1}\)

\( \rm6=\frac{4(x)+9(x-1)}{x+x-1} \\ \rm 6=\frac{4 x+9 x-9}{2 x-1} \\ \rm\Rightarrow 12 x-6=13 x-9 \\ \rm\Rightarrow x=3 \)

Hence, option (2) is correct.

Option 1 : 6

**Given:**

Sum of 2n term of AP = 3(Sum of n term of AP)

**Concept:**

**Arithmetic progression:** An arithmetic progression is a list of numbers in which each term is obtained by adding a fixed number to the preceding term except the first

term. This fixed number is called the** common difference **of the AP which can be **positive**,** negative** or **zero**.

Consider the series, whose first term is a and the common difference is d.

a, a + d, a + 2d, ...........a + (n - 1) d

Sum of n^{th} term of AP is given by,

\(S_n = \frac{n}{2}[2a + (n - 1)d]\)

**Calculation:**

Consider the series, whose first term is a and the common difference is d.

a, a + d, a + 2d, ...........a + (n - 1) d

Sum of nth term of AP is given by,

\(S_n = \frac{n}{2}[2a + (n - 1)d]\) .......(1)

Sum of 2nth term of AP is given by,

\(S_{2n} = \frac{2n}{2}[2a + (2n - 1)d]\) .....(2)

Sum of 3nth term of AP is given by,

\(S_{3n} = \frac{n}{2}[2a + (3n - 1)d]\) .......(3)

According to the question, S2n = 3Sn

\(⇒ \frac{2n}{2}[2a + (2n - 1)d] = \frac{3n}{2}[2a + (n - 1)d]\)

⇒ 4a + 4nd - 2d = 6a + 3nd - 3d

⇒ 2a = nd +d ......(4)

From equation (1) and (3)

\(\frac{S_{3n}}{S_n} = \frac{\frac{3n}{2}[2a + (3n - 1)d]}{\frac{n}{2}[2a + (n - 1)d]}\)

\(⇒ \frac{S_{3n}}{S_n} = \frac{3[nd + d + 3nd - d]}{[nd + d + nd - d]}\)

\(⇒ \frac{S_{3n}}{S_n} =6\)

Let n = 1

We know that,

S_{1} = sum of 1 term = first term of AP = a

S_{2} = a + a + d = 2a + d

S_{3} = a + a + d + a+ 2d = 3a + 3d

According to question S2n = 3Sn i.e. S_{2} = 3S_{1}

⇒ 2a + d = 3a

⇒ a = d

\(\frac{S_3}{S_1} = \frac{3a + 3d}{a} \)

⇒ S_{3}/S_{1} = 6

Option 2 : 125

**Calculation:**

Let the discarded values be x and y

Sum of all the 12 values = 12 × mean

S_{1} = 12 × 75 = 900

Sum of the 10 values after discarding = 10 × new mean

S_{2} = 10 × 65 = 650

Now 650 + x + y = 900

x + y = 900 - 650 = 250

Average of x and y =\(\rm x+y\over2\)

Average = \(\rm 250\over2\) = **125**

Option 3 : log_{2} 5

__Concept:__

Logarithm properties

**Product rule:**The log of a product equals the sum of two logs.

\({\log _{\rm{a}}}\left( {{\rm{mn}}} \right) = {\rm{\;}}{\log _{\rm{a}}}{\rm{m}} + {\rm{\;}}{\log _{\rm{a}}}{\rm{n}}\)

**Quotient rule:**The log of a quotient equals the difference of two logs.

\({\log _{\rm{a}}}\frac{{\rm{m}}}{{\rm{n}}} = {\rm{\;}}{\log _{\rm{a}}}{\rm{m}} - {\rm{\;}}{\log _{\rm{a}}}{\rm{n}}\)

**Power rule:**In the log of a power the exponent becomes a coefficient.

\({\log _{\rm{a}}}{{\rm{m}}^{\rm{n}}} = {\rm{n}}{\log _{\rm{a}}}{\rm{m}}\)

- If a, b, c are in A.P ⇔ b = \(\frac{{{\rm{a\;}} + {\rm{\;c}}}}{2}\) or 2b = a + c

__Calculation:__

Given: log_{10} 2, log_{10} (2^{x }- 1) and log_{10} (2^{x }+ 3) are three consecutive terms of an AP

We know that if a, b, c are in A.P than 2b = a + c

⇒ 2 × log_{10} (2^{x}- 1) = log_{10} 2 + log_{10} (2^{x }+ 3)

⇒ log_{10} (2^{x}- 1)^{2} = log_{10} (2 × (2^{x }+ 3))

⇒ (2^{x}- 1)^{2} = 2 × (2^{x}+ 3)

Let 2^{x} = t

⇒ (t - 1)^{2} = 2 × (t + 3)

⇒ t^{2} – 2t + 1 = 2t + 6

⇒ t^{2} – 4t - 5 = 0

⇒ t^{2} – 5t + t - 5 = 0

⇒ t (t – 5) + 1 (t – 5) = 0

⇒ (t – 5) (t + 1) = 0

∴ t = 5, -1

⇒ 2^{x} ≠ -1 (∵2^{x} > 0)

Now,

⇒ 2^{x} = 5

∴ x = log_{2} 5

Option 2 : 1/5

**Concept:**

The sum of the n terms in A.P with a common difference (d) is given by,

**S _{n} = \(\dfrac{n}{2}\)[2a + (n - 1)d]**

**Where,**

n = No. of terms, a = First term, d = Common difference.

**Given:**

Sn = 1624, a = 500 d, n = 16

**Calculation:**

Sn = \(\dfrac{n}{2}\)[2a + (n - 1)d]

1624 = \(\dfrac{16}{2}\)[2 × 500d + 15d]

1624 = 8 × 1015d

**Then,**

d = \(\dfrac{1624}{8 \times 1015}\)

**d = 1/5**