Sum MCQ Quiz - Objective Question with Answer for Sum - Download Free PDF

Explanation:

Given:

\(\frac{S_p}{S_q} =\frac{p^2}{q^2}\)

⇒ \(\frac{\frac{p}{2}[2a +(p-1)d]}{\frac{q}{2} [2a+(q-1)d]}\) = p²/q²

⇒ \(\frac{2a+(p-1)d}{2a+(q-1)d} =\frac{p}{q}\)

⇒ 2aq +q(p – 1)d = 2ap + p(q – 1)d

⇒ 2a(q – p) = d(q – p) 

⇒ 2a = d or d = 2a

Thus, The common difference is equal to twice of the first term

∴ The Correct answer is Option C

Calculation

Given

\(\mathrm{S}_{20}=\frac{20}{2}[2 \mathrm{a}+19 \mathrm{~d}]=790\)

2a + 19d = 79 …..(1) 

\(S_{10} = \frac{10}{2}[2a+9d]=145\)

⇒ 2a + 9d = 29  …..(2) 

From (1) and (2) a = -8, d = 5 

\(\mathrm{S}_{15}-\mathrm{S}_5=\frac{15}{2}[2 \mathrm{a}+14 \mathrm{~d}]-\frac{5}{2}[2 \mathrm{a}+4 \mathrm{~d}]\)

\(\Rightarrow\frac{15}{2}[-16+70]-\frac{5}{2}[-16+20]\)

⇒ 405 - 10 

⇒ 395 

Hence option (1) is correct

Calculation

\(\frac{\mathrm{n}}{2}(2 \mathrm{a}+(\mathrm{n}-1) 6)=(\mathrm{n}-2) \cdot 180^{\circ}\)

an + 3n² – 3n = (n –2). 180° ...(1)

Now according to question

a + (n –1)6° = 219°

⇒ a = 225° –6n° ...(2)

Putting value of a from equation (2) in (1)

We get

(225n – 6n²) + 3n² – 3n = 180n – 360 

⇒ 2n² – 42n – 360 = 0 

⇒ n² –14n – 120 = 0

n = 20, –6(rejected) 

Calculation

Let a & d are first term and common difference of an AP. 

\(\mathrm{S}_{40}=\frac{40}{2}[2 \mathrm{a}+39 \mathrm{~d}]=1030 ...(1)\)

\(\mathrm{S}_{12}=\frac{12}{2}[2 \mathrm{a}+11 \mathrm{~d}]=57 ...(2)\)

by (1) & (2) 

\(\mathrm{a}=-\frac{7}{2} \ and\ \mathrm{~d}=\frac{3}{2}\)

∴ \(\mathrm{S}_{30}-\mathrm{S}_{10}=\frac{30}{2}[2 \mathrm{a}+29 \mathrm{~d}]-\frac{10}{2}[2 \mathrm{a}+9 \mathrm{~d}]\)

= 20a – 390d

= 515 

Hence option 2 is correct

Let a₁, a₂ be the first terms of the two progressions.

Let d₁, d₂ be the common differences of the two progression.

We need to find \(\frac{\mathrm{a}_{1}+10 \mathrm{~d}_{1}}{\mathrm{a}_{2}+10 \mathrm{~d}_{2}}\)

Given \(\frac{\frac{\mathrm{n}}{2}\left[2 \mathrm{a}_{1}+(\mathrm{n}-1) \mathrm{d}_{1}\right]}{\frac{\mathrm{n}}{2}\left[2 \mathrm{a}_{2}+(\mathrm{n}-1) \mathrm{d}_{2}\right]}=\frac{7 \mathrm{n}+4}{6 \mathrm{n}-5}\)

\(=\frac{2 \mathrm{a}_{1}+(\mathrm{n}-1) \mathrm{d}_{1}}{2 \mathrm{a}_{2}+(\mathrm{n}-1) \mathrm{d}_{2}}=\frac{7 \mathrm{n}+4}{6 \mathrm{n}-5}\)

Therefore, \(\frac{2 a_{1}+0 d_{1}}{2 a_{2}+20 d_{2}}=\frac{7(21)+4}{6(21)-5} \Rightarrow \frac{a_{1}+10 d_{1}}{a_{2}+10 d_{2}}=\frac{151}{121}\)

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)

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:

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:

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

Concept:

The nth number in A.P. series (a_n) = a + (n-1)d

The sum of the n numbers in the series = \(\rm {n\over2}\left[2a + (n-1)d\right]\)

Where 'a' is the first number of the series and 'd' is the common difference

Calculation:

Let the first term of the A.P. be 'a' and the common difference be 'd'

Given a₃ + a₇ = 30

a + 2d + a + 6d = 30

2a + 8d = 30    ...(i)

Also given a₅ + a₉ = 56 

a + 4d + a + 8d = 56

2a + 12d = 56   ...(ii)

Adding (i) and (ii)

4a + 20d = 86

2a + 10d = 43

a + 3d + a + 7d = 43

a₄ + a₈ = 43

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:

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 = 14n + 3

T₁ = a =  14(1) + 3 = 17

Sum of n^th term S_n

⇒ \(\rm n\over 2\) (17 + 14n + 3)

⇒ \(\rm n\over 2\)(14n + 20) 

⇒ 7n² + 10n

Alternate Method

An arithmetic progression (AP) is defined by a constant difference between consee terms, and its nth term can be expressed as an = a + (n - 1)d where a is the t and d is a common difference.

You've been given that the nth term is 14n + 3 and we want to find the sum of the first n terms.

First, we can find the first term and the common difference by plugging in the value n = 1 n = 2

a1 = 14(1) + 3 = 17

a2 = 14(2) + 3 = 31

So, we can find the common difference:

d = a2 - a1 = 31 - 17 = 14

Now, we can use the formula for the sum of the first n terms of an AP:

\(S_n=\frac n2(2a+(n-1)d)\)

\(S_n=\frac n2(2 \times 17+(n-1)14)\)

\(S_n=\frac n2(34+14n-14)\)

\(S_n=\frac n2(20+14n)\)

Sn = 7n2 + 10n

Shortcut Trick

Take n = 2

T1 = first term(a) =  14(1) + 3 = 17

T₂ = second term = 14(2) + 3 = 28 + 3 = 31

Sum of 2 term = 17 + 31 = 48

Check option

(1) 10n² + 7n = 10(2)² + 7(2) = 10(4) + 14 = 54

(2) 7n² + 8n = 7(2)² + 8(2) = 28 + 16 = 44

(3) 7n² + 10n = 7(2)² + 10(2) = 28 + 20 = 48

Then option (3) is correct

Hint

In A.P. question you will check the option first than take the value of n