Sum MCQ Quiz - Objective Question with Answer for Sum - Download Free PDF
Last updated on Apr 11, 2025
Latest Sum MCQ Objective Questions
Sum Question 1:
In an AP, the ratio of the sum of the first p terms to the sum of the first q terms is p2 : q2. Which one of the following is correct?
Answer (Detailed Solution Below)
Sum Question 1 Detailed Solution
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
Sum Question 2:
Let Sa denote the sum of the first n terms in an arithmetic progression. If S20 = 790 and S10 = 145, then S15 – S5 is:
Answer (Detailed Solution Below)
Sum Question 2 Detailed Solution
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
Sum Question 3:
The interior angles of a polygon with n sides, are in an A.P. with common difference 6°. If the largest interior angle of the polygon is 219°, then n is equal to _____.
Answer (Detailed Solution Below) 20
Sum Question 3 Detailed Solution
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)
Sum Question 4:
In an arithmetic progression, if S40 = 1030 and S12 = 57, then S30 – S10 is equal to:
Answer (Detailed Solution Below)
Sum Question 4 Detailed Solution
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
Sum Question 5:
Find the ratio of the 11th term of the two arithmetic progression if the ratio of the sum up to the two progressions is (7n + 4) : (6n - 5).
Answer (Detailed Solution Below)
Sum Question 5 Detailed Solution
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, \(\frac{2 a_{1}+0 d_{1}}{2 a_{2}+20 d_{2}}=\frac{7(21)+4}{6(21)-5} \Rightarrow \frac{a_{1}+10 d_{1}}{a_{2}+10 d_{2}}=\frac{151}{121}\)
Top Sum MCQ Objective Questions
The sum of the series 5 + 9 + 13 + … + 49 is:
Answer (Detailed Solution Below)
Sum Question 6 Detailed Solution
Download Solution PDFConcept:
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.
Find the sum to n terms of the A.P., whose nth term is 5n + 1
Answer (Detailed Solution Below)
Sum Question 7 Detailed Solution
Download Solution PDFConcept:
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 an = 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)
The tenth term common to both the A. P. 3, 7, 11, ... and 1, 6, 11, ... is:
Answer (Detailed Solution Below)
Sum Question 8 Detailed Solution
Download Solution PDFConcept:
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.
The sum of (p + q)th and (p – q)th terms of an AP is equal to
Answer (Detailed Solution Below)
Sum Question 9 Detailed Solution
Download Solution PDFConcept:
The nth term of an AP is given by: Tn = 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: Tn = 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
\(\Rightarrow \;{a_{p + q}} + {a_{p - q}} = 2\;a + 2\;\left( {p - 1} \right)d = 2 \times \left[ {a + \left( {p - 1} \right)d} \right] = 2 \times {a_p}\)Find the sum of all numbers divisible by 6 in between 100 to 400
Answer (Detailed Solution Below)
Sum Question 10 Detailed Solution
Download Solution PDFConcept:
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
nth term = 396
As we know, nth term of AP = an = 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
If the first term of an AP is 2 and the sum of the first five terms is equal to one-fourth of the sum of the next five terms, then what is the sum of the first ten terms?
Answer (Detailed Solution Below)
Sum Question 11 Detailed Solution
Download Solution PDFConcept:
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.= an = 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,
an = nth 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.
What is the sum of first n odd natural numbers?
Answer (Detailed Solution Below)
Sum Question 12 Detailed Solution
Download Solution PDFConcept:
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
The sum of the 3rd and the 7th term of an A.P. is 30 and the sum of the 5th and the 9th term is 56. Find the sum of the 4th and the 8th terms of the same series.
Answer (Detailed Solution Below)
Sum Question 13 Detailed Solution
Download Solution PDFConcept:
The nth number in A.P. series (an) = 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
Find the sum of all the numbers divisible by 6 in between 100 to 400
Answer (Detailed Solution Below)
Sum Question 14 Detailed Solution
Download Solution PDFConcept:
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
nth term = 396
As we know, nth term of AP = an = 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
If the nth term of A.P. is 14n + 3, so find the sum of nth term.
Answer (Detailed Solution Below)
Sum Question 15 Detailed Solution
Download Solution PDFConcept:
nth term of A.P., Tn = a + (n - 1)d
Sum of nth term of A.P., Sn = \(\rm n\over 2\)[2a + (n - 1)d] = \(\rm n\over 2\)(a + Tn)
Here, a = firse term, Tn = nth term (last term)
Calculaion:
Tn = 14n + 3
T1 = a = 14(1) + 3 = 17
Sum of nth term Sn
⇒ \(\rm n\over 2\) (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