Arithmetic Progressions MCQ Quiz - Objective Question with Answer for Arithmetic Progressions - Download Free PDF

Given:

The sum of n terms of an A.P. is given by:

S_n = 3n² + n

The common difference, d = 6

Formula used:

The sum of the first n terms of an A.P. is given by the formula:

S_n = n/2 × (2a + (n - 1)d)

Where, a = first term, d = common difference

Calculations:

We are given S_n = 3n² + n, and d = 6. We can equate this to the formula for the sum of n terms of an A.P.:

3n² + n = n/2 × (2a + (n - 1) × 6)

Multiply both sides by 2 to eliminate the fraction:

6n² + 2n = n × (2a + 6n - 6)

Now, divide both sides by n (n ≠ 0):

6n + 2 = 2a + 6n - 6

Cancel out 6n from both sides:

2 = 2a - 6

2a = 8

a = 4

∴ The first term of the A.P. is 4.

Concept:

In an AP, the nth term is given by:

T_n = a + (n - 1)d

In a GP, terms satisfy the relation:

T_m² = T_n × T_p

Calculation:

Let the 5th, 7th, and 13th terms of the AP be T₅, T₇, and T₁₃ respectively.

⇒ T₅ = a + 4d

⇒ T₇ = a + 6d

⇒ T₁₃ = a + 12d

Since the terms are in GP:

⇒ T₇² = T₅ × T₁₃

⇒ (a + 6d)² = (a + 4d) × (a + 12d)

Expanding both sides:

⇒ (a + 6d)² = a² + 12ad + 48d²

⇒ a² + 12ad + 36d² = a² + 12ad + 48d²

Cancel out common terms:

⇒ 36d² = 48d²

⇒ -12d² = 0

Dividing through by d² (assuming d ≠ 0):

⇒ a = -3d

Conclusion:

∴ Ratio of the first term to the common difference is:

⇒ a/d = -3

Hence, the correct answer is Option 1.

Given:

The sum of the first k terms of a series S is given as S_k = 3k² + 5k.

Concept:

To find the nth term of a series, we use the formula:

a_n = S_n - S_n-1

If the nth term forms an arithmetic progression (AP), the common difference (d) is given by:

d = a_n+1 - a_n

Calculation:

We are given S_k = 3k² + 5k.

⇒ a_n = S_n - S_n-1

Also S_n = 3n² + 5n and S_n-1 = 3(n-1)² + 5(n-1)

⇒ a_n = [3n² + 5n] - [3(n-1)² + 5(n-1)]

⇒ a_n = [3n² + 5n] - [3(n² - 2n + 1) + 5n - 5]

⇒ a_n = [3n² + 5n] - [3n² - 6n + 3 + 5n - 5]

⇒ a_n = 3n² + 5n - 3n² + 6n - 3 - 5n + 5

⇒ a_n = 6n + 2

The series is an arithmetic progression (AP) if the difference between consecutive terms is constant.

⇒ Common difference d = a_n+1 - a_n

⇒ a_n+1 = 6(n+1) + 2 = 6n + 6 + 2 = 6n + 8

⇒ d = (6n + 8) - (6n + 2) = 6

∴ The terms of the series form an arithmetic progression with a common difference of 6.

Hence, the correct answer is Option B.

Answer (8)

Sol.

Let the number of terms be 2n 

T₁ + T₃ + T₅ ... T_2n-1 = 24

\(\rm \frac{T_{2}+T_{4}+T_{6} \ldots T_{2 n}=30}{\left.T_{2}-T_{1}\right)+\left(T_{4}-T_{3}\right)+\ldots\left(T_{2 n}-T_{2 n-1}\right)=6}\)

nd = 6

(a + (2n + 1)d) - a = \(\frac{21}{2}\)

⇒ 2nd - d = \(\frac{21}{2}\)

⇒ 12 - \(\frac{21}{2}\) = d

⇒ d = \(\frac{3}{2}\)

∴ n = 4

∴ Total terms = 8

Concept:

• Arithmetic Progression (A.P.): The n^th term of an A.P. is given by: T_n = a + (n − 1)d
• Determinant of 3×3 matrix: For matrix: , the determinant is
• Sum of A.P.: The sum of first n terms is S_n = (n/2)[2a + (n − 1)d]

Calculation:

Let A₁, A₂, A₃ be three A.P.s with first terms: A, A + 1, A + 2 respectively and common difference = d

⇒ 7^th term of A₁:

⇒ 9^th term of A₂:

⇒ 17^th term of A₃:

Given determinant condition:

\(\left|\begin{array}{lll} \mathrm{a} & 7 & 1 \\ 2 \mathrm{~b} & 17 & 1 \\ \mathrm{c} & 17 & 1 \end{array}\right|+70=0\)

Expand determinant along row 1:

= a(17×1 − 1×17) − 7(2b×1 − 1×c) + 1(2b×17 − 17×c)

= a(0) − 7(2b − c) + 1(34b − 17c) + 70 = 0

⇒ −14b + 7c + 34b − 17c + 70 = 0

⇒ (20b − 10c + 70 = 0)

⇒ 2b − c = −7

Now, given: a = A + 6d = 29

⇒ A = 29 − 6d

Also: b = A + 1 + 8d = (29 − 6d) + 1 + 8d = 30 + 2d

c = A + 2 + 16d = (29 − 6d) + 2 + 16d = 31 + 10d

Now check condition: 2b − c = −7

2(30 + 2d) − (31 + 10d) = 60 + 4d − 31 − 10d = 29 − 6d = −7

⇒ 29 + (−6d) = −7

⇒ −6d = −36

⇒ d = 6

Now find A = 29 − 6×6 = −7

Then b = 30 + 2×6 = 42

c = 31 + 10×6 = 91

Now required:

First term = c − a − b = 91 − 29 − 42 = 20

Common difference = d / 12 = 6 / 12 = 0.5

Sum of first 20 terms:

S20 = (20 / 2)[2×20 + (20 − 1)×0.5]

= 10[40 + 9.5]

= 10 × 49.5

= 495

∴ The required sum is 495.

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:
  Sn = \(\rm \dfrac{n}{2}[a+\{a+(n-1)d\}]=\left (\dfrac{First\ Term+Last\ Term}{2} \right )\times n\)

Calculation:

The given series is 5 + 9 + 13 + … + 49 which is an arithmetic progression with first term a = 5 and common difference d = 4.

Let's say that the last term 49 is the nth term.

∴ a + (n - 1)d = 49

⇒ 5 + 4(n - 1) = 49

⇒ 4(n - 1) = 44

⇒ n = 12.

And, the sum of this AP is:

S₁₂ = \(\rm \left (\dfrac{First\ Term+Last\ Term}{2} \right )\times 12\)

= \(\rm \left (\dfrac{5+49}{2} \right )\times 12\) = 54 × 6 = 324.

Concept:

For AP series, 

Sum of n terms  = \(\rm \dfrac n 2\) (First term + nth term)

Calculations:

We know that, For AP series, 

the sum of n terms  = \(\rm \dfrac n 2\) (First term + nth term)

Given, the nth term of the given series is a_n = 5n + 1.

Put n = 1, we get

a1 = 5(1) + 1 = 6.

We know that 

sum of n terms = \(\rm \dfrac n 2\) (First term + nth term)

⇒Sum of n terms = \(\rm \dfrac n 2\) (6 + 5n + 1)

⇒Sum of n terms = \(\rm \dfrac n 2\) (7+ 5n)

GIVEN:

Two series given i.e. S1 and S2

FORMULA USED:

a_n = a + ( n - 1 ) d 

S_n = n/2 [2a + (n - 1) d ] 

Where,

a_n = n^th term in the sequence , n= number of terms , a = first term in sequence, d = common difference , Sn = Sum 

CALCULATION:

Here, given series S1  and S2 are in A.P.

So, series will move by adding a fixed common difference ( second term - first term) in consecutive terms 

S1 = 2 , 9 , 16 , 23, 30 , 37 , 44 , 51 ,........ 632      [As  here d = 7 .So, add 7 in previous term to get next term]

S2  = 7, 11, 15, 19, 23, 27, 31, 35, 39, 43, 47, 51 , ......... 743  [ Here d = 4 ] 

Now, let us take a third series S3 which is common series                    [It will contain common numbers of both series only]

So, from S1 and S2 series, we have 1st common term = 23 , 2nd common term = 51 , d = ( 51 - 23) = 28 

So, add 28 in second term to get third term and so on

S3 = 23 , 51 , .................. \(\le\)  632              [ As, 632 is less than 743 so common between them should be less than 632]

Now, we have a = 23 , d = 28 

⇒ a_n = a + ( n - 1) d    \(\le\) 632 

⇒ 23 + ( n - 1) × 28 \(\le\) 632 

⇒ ( n -  1) × 28  \(\le\)  ( 632 - 23 ) 

⇒ ( n - 1) × 28  \(\le\) 609 

⇒ n -1 \(\le\) 609 /28 

⇒ n - 1 \(\le\) 21.75 

⇒ n \(\le\) 22 .75

 As, n should be equal to or less than 22.75 . So, take n = 22 

Now, as we know that 

Sn = n/2 [ 2a + ( n - 1) d ] 

⇒ Sn = 22/2 [ 2 × 23 + ( 22 - 1) 28]  = 11 [46 + 21 × 28 ] 

⇒ 11 [ 46 + 588 ] = 11 × 634  = 6974

Hence, Sum of all the common terms of the series are 6974 .

Concept:

Arithmetic Progressions:

• The series of numbers where the difference of any two consecutive terms is the 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

• Common Terms to two A. P.s form an A. P. themselves, with common difference equal to the LCM of the common difference of the two A. P.s.

Calculation:

For the given two A. P.s 3, 7, 11, ... and 1, 6, 11, ..., the common differences are 4 and 5 respectively and 11 is the first common term.

The common difference of the terms common to both the series will be: LCM of (4 and 5) = 20.

The required 10th term common to both the A. P.s = a + (n - 1)d

= 11 + (10 - 1) × 20

= 11 + 180

= 191.

Concept:

If a, b, c are in A.P then 2b = a + c

Calculation:

Given:

n - 3, 4n - 2, 5n + 1 are in AP

Therefore, 2 × (4n - 2) = (n - 3) + (5n + 1)

⇒ 8n - 4 = 6n - 2

⇒ 2n = 2

∴ n = 1

Concept:

The nth term of an AP is given by: T_n = a + (n - 1) × d, where a = first term and d = common difference.

Calculation:

As we know that, the nth term of an AP is given by: T_n = a + (n - 1) × d, where a = first term and d = common difference.

Let a be the first term and d is the common difference.

\(\Rightarrow \;{a_{p + q}} = a + \left( {p + q - 1} \right) \times d\)     ...1)

\(\Rightarrow \;{a_{p - q}} = a + \left( {p - q - 1} \right) \times d\)     ...2)

By adding (1) and (2), we get

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:

Here 1st term = a = 102 (Which is the 1st term greater than 100 that is divisible by 6.) 

The last term less than 400, Which is divisible by 6 is 396. 

Terms in the AP; 102, 108, 114 … 396

Now

First term = a = 102

Common difference = d = 108 - 102 = 6

n^th term = 396

As we know, n^th term of AP = a_n = a + (n – 1) d

⇒ 396 = 102 + (n - 1) × 6

⇒ 294 = (n - 1) × 6

⇒ (n - 1) = 49

∴ n = 50

Now,

Sum =  \(\rm \frac n 2\)(a + l) =  \(\rm \frac {50}{2}\)(102 + 396) = 25 × 498 = 12450

Concept:

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

Common difference “d”= a₂ – a₁ = a₃ – a₂ = …. = a_n – a_{n – 1}

n^th term of the A.P.= a_n = a + (n – 1) d

Sum of the first n terms (S) =  \(\frac{n}{2}[2a+(n-1)d)]\)

Also, S =  (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, a₁ = 2      ...(1)

S5 = \(1\over4\)(S10 - S5)

⇒ 4S5 = S10 - S5

⇒ 4S5 + S5 = S10

⇒ 5S5 = S10

⇒ 5 × \(5\over2\)[2a1 + (n5 - 1)d] = \(10 \over2\) [2a1 + (n10 - 1)d]

[∵ n5 = 5 और n10 = 10]

⇒ 5 × \(5\over2\)[a1 + a1 + 4d] = \(10 \over2\) [a1 + a1 + 9d]

⇒ 5 × [2a1 + 4d] = 2 × [2a1 + 9d]

⇒ 10a1 + 20d = 4a1 + 18d

⇒ 6a1 = 2d

⇒ a1 = \(\rm-d\over3\)

∴ d = -3a1 = - 3 × 2 = - 6

Hence,

S10 = \(10 \over2\) [a1 + a1 + 9d]

⇒ 5[2a1 + 9d]

⇒ 5[4 - 54] = -250

∴ S10 = 5 × (-50) = - 250.

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: Fourth term of an A.P. is zero

⇒ a₄ = 0

⇒ a + (4 - 1) × d = 0

⇒ a + 3d = 0         

∴ a = -3d                     .... (1)

To Find: \(\rm \dfrac{t_{25}}{t_{11}}\)

\(\rm \Rightarrow \dfrac{t_{25}}{t_{11}} = \dfrac {a+24d}{a+10d}\\=\dfrac {-3d+24d}{-3d+10d}\\=\dfrac {21d} {7d}=3\)

Concept:

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

• Common difference “d”= a₂ – a₁ = a₃ – a₂ = …. = a_n – a_{n – 1}

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

Calculation:

Given series is 2, 6, 10, ...,146

First term, a = 2, last term, a_n = 146, an

Common difference d = 4, so it is an AP

an = a + (n – 1) d

146 = 2 + (n - 1) (4)

⇒ n - 1 = 144/4

⇒ n = 36 + 1 = 37

So, number of terms in given series = 37

Middle term = (37 + 1)/2 = 19th term 

a₁₉ = 2 + (19 - 1) × 4

= 2 + 72 

= 74

Hence, option (3) is correct.