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Option 3 : 1

**Given: **

For a value of k; 2, 3 + k and 6 are to be in A.P

**Concept:**

According to Arithmetic progression, a2 - a1 = a3 - a2

where a1 ,a2 ,a3 are 1st, 2nd and 3rd term of any A.P.

**Calculation:**

a_{1} = 2, a_{2 }= k + 3, a_{3} = 6 are three consecutive terms of an A.P.

According to Arithmetic progression, a_{2} - a_{1 }= a_{3} - a_{2 }

(k + 3) – 2 = 6 – (k + 3)

⇒ k + 3 - 8 + k + 3 = 0

⇒ 2k = 2

After solving, we get k = 1

Option 4 : 2b = a + c

**Concept used:**

Let a, b, c… and so on be our series

As we know the common difference = b – a, c – b.

The common difference is the same in arithmetic progression

b – a = c – b

**Calculation:**

b – a = c – b

⇒ b + b = c + a

⇒ 2b = c + a

⇒ 2b = a + c

**∴ a, b, c are in arithmetic progression then 2b = a + c.**

__Alternate Method__

Let number be 1, 2, 3 which are in AP

Only one option satisfied the equation

2(2 ) 1 + 3 =so 2b = a + c is correct option

What will come in place of the question mark (?) in the following question?

13 + 23 + 33 + ……+ 93 = ?

Option 1 : 477

**Given****:**

13 + 23 + …….. + 93 = ?

**Formula:**

S_{n} = n/2 [a + l]

T_{n} = a + (n – 1)d

n = number of term

a = first term

d = common difference

l = last term

**Calculation:**

a = 13

d = 23 – 13 = 10

T_{n} = [a + (n – 1)d]

⇒ 93 = 13 + (n – 1) × 10

⇒ (n – 1) × 10 = 93 – 13

⇒ (n – 1) = 80/10

⇒ n = 8 + 1

⇒ n = 9

S_{9} = 9/2 × [13 + 93]

= 9/2 × 106

= 9 × 53

= 477

Option 2 : 551

**GIVEN:**

The sum of 6 consecutive odd numbers is 144.

**CALCULATION:**

Let six consecutive odd numbers be x, x + 2, x + 4, x + 6, x + 8 and x + 10.

Now,

x + x + 2 + x + 4 + x + 6 + x + 8 + x + 10 = 144

⇒ 6x + 30 = 144

⇒ 6x = 144 – 30

⇒ x = 114/6

⇒ x = 19

∴ x + 10 = 29

Hence,

Product of the first number and the last number = 19 × 29 = 551Option 2 : 81

**GIVEN:**

Sum of 4^{th}, 7^{th} and 11^{th} term of an arithmetic progression is 203.

Common difference = 8

**CONCEPT: **

Arithmetic progression formulas for calculating n^{th} term.

**FORMULA USED:**

n^{th} term of an arithmetic progression = [a + (n – 1) × d]

Where

a = first term, d = common difference, n = number of terms

**CALCULATION:**

Using the formula,

⇒ [a + (4 – 1)d + a + (7 – 1)d + a + (11 – 1)d = 203

⇒ 3a + 19d = 203

As given, d = 8

⇒ 3a = 203 – 152

⇒ a = 17

⇒ 9^{th} term = 17 + (9 – 1) × 8 = 81

∴ the 9^{th} term of the progression is 81.

Option 4 : 340

**GIVEN:**

AP of 10 terms having 1^{st} term as 7 and last term as 61

**FORMULA USED:**

a_{n} = a + (n - 1) × d

S_{n} = n/2(2a + (n - 1) × d)

**CALCULATION:**

a = 7

And, a_{10} = 61

⇒ a + 9d = 61

⇒ 7 + 9d = 61

⇒ d = 6

Now sum of all 10 terms = 10/2(2 × 7 + 9 × 6) = 340Option 1 : 17.5

**GIVEN:**

First term = 7

Third term = 28

**CONCEPT:**

When three quantities are in AP, the middle one is called as the arithmetic mean of the other two.

**FORMULAE USED:**

If a, b and c are three terms in AP

Then b = (a + c)/2

**CALCULATION:**

Second term = (7 + 28)/2 = 35/2

∴ Second term = 17.5

__Alternate Method__

a = 7,

a + 2d = 28

⇒ 2d = 28 - 7 = 21

⇒ d = 10.5

2nd term = a + d = 7 + 10.5 = 17.5

Option 4 : 395

**GIVEN:**

First odd number is 1, second is 3, third is 5 and fourth is 7

**FORMULA USED:**

a_{n} = a + (n - 1) × d

S_{n} = n/2[2a + (n - 1) × d]

Where n = number of terms, a = first term, a_{n} = n^{th} term, d = common difference, and S_{n} = sum of arithmetic progression

**CALCULATION:**

a = 1

a_{2} = 3

⇒ a + d = 3

⇒ 1 + d = 3

⇒ d = 2

Now, a_{198} = a + 197d = 1 + 197 × 2 = 395

**Thus the correct answer is 395.**

Option 1 : 15

**Given,**

2^{nd} term of AP = 17

11^{th} term of AP = 35

**Formula:**

T_{n} = a + (n – 1)d

Where, a = first term and d = common difference

**Calculation:**

T_{2} = a + (n – 1) × d

⇒ 17 = a + d ----(1)

T_{11} = a + (n – 1) × d

⇒ 35 = a + 10d ----(2)

Subtract equation (1) from equation (2)

18 = 9d

⇒ d = 18/9

⇒ d = 2

Putting d = 3 in equation (1), we get

17 = a + 2

⇒ a = 17 – 2

⇒ a = 15

**∴ First term is 15.**

Option 1 : 160

**Given:**

First term is 7 and 15^{th} term is 35

**Formula used:**

In an arithmetic progression, nth term = a + (n – 1) d, where a = first term and d = common difference

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

**Calculation:**

a = 7,

15^{th} term = a + 14 d = 35

⇒ 7 + 14 d = 35

⇒ 14 d = 28

⇒ d = 2

∴ Sum of the first 10 terms = 10/2 × [2a + 9 d]

= 5 × [14 + 18]

= 160

Option 3 : 465

**Formula:**

Sum of first n natural numbers = [n (n + 1)]/2

**Calculation:**

n = 30

Sum of first 30 natural numbers = [30 × 31]/2 = 465Option 4 : 19

**Given,**

N^{th} term of AP = 0

The given AP = 54, 51, 48 …

**Formula:**

T_{n} = a + (n – 1)d

Where, a = first term and d = common difference

**Calculation:**

Here,

a = 54

d = 51 – 54 = - 3

T_{n} = a + (n – 1) d

⇒ 0 = 54 + (n – 1) × (- 3)

⇒ (n – 1) × (-3) = - 54

⇒ n – 1 = 54/3

⇒ n – 1 = 18

⇒ n = 19

∴ There are 19 terms of the AP.

Option 4 : - 196

**GIVEN:**

First term = 25

Last term = -53

N = 14

**FORMULAE USED:**

Sum of first n terms = N × (first term + last term)/2

**CALCULATION:**

Sum = 14/2 (25 - 53) = 7 × (-28)

Option 2 : 9 : 11

**Given: **

In an AP, the ratio of the fifth term to the eleventh term is 2 : 3. The sixth term is 13.

**Formula used: **

nth term of an AP, T_{n }= a + (n - 1)d,

where ‘a’ is the first term, ‘n’ is the total number of terms and ‘d’ is the common difference.

**Calculation:**

Let the first term is ‘a’ and common difference is ‘d’.

According to question

T_{5}/T_{11} = 2/3

⇒ (a + 4d)/(a + 10d) = 2/3

⇒ 3a + 12d = 2a + 20d

⇒ a – 8d = 0 ----(1)

And T_{6 }= 13 (given)

⇒ a + 5d = 13 ----(2)

Solving equation (1) and (2)

8d + 5d = 13

⇒ 13d = 13

⇒ d = 1

and by putting value d = 1 in equation (1) we get,

a = 8

The ratio of the eleventh term and the fifteenth term = (a + 10d)/(a + 14d)

⇒ (8 + 10 × 1)/(8 + 14 × 1)

⇒ 18/22

**∴ required answer is 9/11.**

Find the value of 6 + 11 + 16 + 21 + … + 71.

A. 539

B. 561

C. 661

D. 639

Option 1 : A

**Given****:**

The given series is 6 + 11 + 16 + 21 + … + 71.

**Formula:**

T_{n} = a + (n – 1) d

S_{n} = (n/2) [2a + (n – 1) d]

a = first term

d = common difference

**Calculation:**

a = 6

d = 11 – 6 = 5

⇒ 71 = 6 + (n – 1) 5

⇒ (n – 1) 5 = 71 – 6

⇒ n – 1 = 65/5

⇒ n = 13 + 1

⇒ n = 14

∴ S_{14} = (14/2) × [2 × 6 + (14 – 1) × 5]

= 7 × [12 + 65]

= 7 × 77 = 539

Option 2 : 23

**Given:**

The average of 16, 17, 18, ..... up to fifteen terms

n = 15, a = 16, d = 1

**Formula used:**

For an A.P of n numbers with common difference d and first term a

Sum of n numbers = (n/2) × (2a + (n – 1 ) × d)

Average = Sum/number of terms

**Calculation:**

Here, a = 16, d = 1 and n = 15

According to the question,

Sum of n numbers = (n/2) × (2a + (n – 1 ) × d)

⇒ Sum = (15/2) × ((2 × 16) + (15 – 1) × 1))

⇒ 345

Average = 345/15

⇒ 23

**∴ The average of 16, 17, 18, ..... up to fifteen terms will be 23.**

Option 3 : 52

**Given**

Average of 5 consecutive numbers is 26.

**Formula**

Sum of numbers = Average × Total numbers

**Calculation**

Lets numbers be a, a + 1, a + 2, a + 3 and a + 4.

According to question

a + (a + 1) + (a + 2) + (a + 3) + (a + 4) = 26 × 5

⇒ 5a + 10 = 130

⇒ 5a = 120

⇒ a = 24

Hence,

First number = a = 24

Fifth number = a + 4 = 24 + 4 = 28

∴ Sum = 24 + 28 = 52.

Option 1 : 513

**GIVEN:**

7^{th} term is 21 and 12^{th} term is 36

**FORMULA USED:**

a_{n} = a + (n - 1) × d

S_{n} = n/2(2a + (n - 1) × d)

**CALCULATION:**

a_{7} = 21

⇒ a + 6d = 21 ……(i)

Also, a_{12} = 36

⇒ a + 11d = 36 ……(ii)

On equating both the equations we get,

a = 3 and d = 3

∴ S_{18} = 18/2(2 × 3 + 17 × 3) = 513

Option 2 : 5775

**Formula Used**:

sum of first n terms = \(\frac{{n\left( {a + l} \right)}}{2}\)

**Calculation**:

The above given question can also be written as,

⇒ (90 + 1)(90 + 2)(90 + 3) ....... + (90 + 50)

⇒ (90 + 90 + 90 .... 50 terms) + (1 + 2 + 3 + ... + 50)

⇒ 4500 + \(\frac{{n\left( {a + l} \right)}}{2}\)

⇒ where n = 50 because we have 50 terms,

⇒ 4500 + 25 ( 51 )

⇒ 5775

Option 3 : 6

**Formula Used**:

a_{n} = a + (n - 1)d; where a = first term, d = common difference, n = number of term, a_{n} = n^{th} term

**Calculation:**

200 = 150 + (n - 1) × 9

⇒ 50 = (n - 1) × 9

⇒ (n - 1) = 5.55

⇒ n = 6.55 ≈ 6

**∴** Required number = 6