Sum MCQ Quiz - Objective Question with Answer for Sum - Download Free PDF
Last updated on Apr 25, 2025
Latest Sum MCQ Objective Questions
Sum Question 1:
Find the sum of 3 +32 + 33 +...+ 38.
Answer (Detailed Solution Below)
Sum Question 1 Detailed Solution
Formula used:
Sum of geometric progression (Sn) = {a × (rn - 1)}/(r - 1)
Where, a = first term ; r = common ratio ; n = number of term
Calculation:
3 +32 + 33 +...+ 38.
Here, a = 3 ; r = 3 ; n = 8
Sum of the series (S8) = {a × (r8 - 1)}/(r - 1)
⇒ {3 × (38 - 1)}/(3 - 1)
⇒ (3 × 6560)/2 = 3280 × 3
⇒ 9840
∴ The correct answer is 9840.
Sum Question 2:
The nth terms of the two series 3 + 10 + 17 + ... and 63 + 65 + 67 + .... are equal, then the value of n is:
Answer (Detailed Solution Below)
Sum Question 2 Detailed Solution
Given:
nth term for the first series = nth term for the second series.
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 and it can be positive, negative or zero.
Consider the AP, whose first term is 'a' and common difference is 'd' and there are n number of terms.
a, a + d, a + 2d, a + 3d........a + (n - 1)d
Hence, nth term of AP is given by
Tn = a + (n - 1)d
Calculation:
Consider the first series
3 + 10 + 17 + ...
a = 3 and d = 10 - 3 = 7
nth term of AP is given by,
(Tn)1 = 3 + (n - 1) × 7
⇒ (Tn)1 = 7n - 4 .......(1)
Consider the second series
63 + 65 + 67 + ....
a = 63 and d = 2
nth term of AP is given by,
(Tn)2 = 63 + (n - 1) × 2
⇒ (Tn)2 = 2n + 61 .......(2)
According to the question
(Tn)1 = (Tn)2
⇒ 7n - 4 = 2n + 61
⇒ 5n = 65
⇒ n = 13
Hence, for the given series, 13th term will be equal.
Additional Information
Sum of n terms of series, whose first term is a and the common difference is d.
\(S_n = \frac{n}{2}[ 2a + (n - 1)d]\)
Top Sum MCQ Objective Questions
Find the sum of 3 +32 + 33 +...+ 38.
Answer (Detailed Solution Below)
Sum Question 3 Detailed Solution
Download Solution PDFFormula used:
Sum of geometric progression (Sn) = {a × (rn - 1)}/(r - 1)
Where, a = first term ; r = common ratio ; n = number of term
Calculation:
3 +32 + 33 +...+ 38.
Here, a = 3 ; r = 3 ; n = 8
Sum of the series (S8) = {a × (r8 - 1)}/(r - 1)
⇒ {3 × (38 - 1)}/(3 - 1)
⇒ (3 × 6560)/2 = 3280 × 3
⇒ 9840
∴ The correct answer is 9840.
The nth terms of the two series 3 + 10 + 17 + ... and 63 + 65 + 67 + .... are equal, then the value of n is:
Answer (Detailed Solution Below)
Sum Question 4 Detailed Solution
Download Solution PDFGiven:
nth term for the first series = nth term for the second series.
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 and it can be positive, negative or zero.
Consider the AP, whose first term is 'a' and common difference is 'd' and there are n number of terms.
a, a + d, a + 2d, a + 3d........a + (n - 1)d
Hence, nth term of AP is given by
Tn = a + (n - 1)d
Calculation:
Consider the first series
3 + 10 + 17 + ...
a = 3 and d = 10 - 3 = 7
nth term of AP is given by,
(Tn)1 = 3 + (n - 1) × 7
⇒ (Tn)1 = 7n - 4 .......(1)
Consider the second series
63 + 65 + 67 + ....
a = 63 and d = 2
nth term of AP is given by,
(Tn)2 = 63 + (n - 1) × 2
⇒ (Tn)2 = 2n + 61 .......(2)
According to the question
(Tn)1 = (Tn)2
⇒ 7n - 4 = 2n + 61
⇒ 5n = 65
⇒ n = 13
Hence, for the given series, 13th term will be equal.
Additional Information
Sum of n terms of series, whose first term is a and the common difference is d.
\(S_n = \frac{n}{2}[ 2a + (n - 1)d]\)
Sum Question 5:
Find the sum of 3 +32 + 33 +...+ 38.
Answer (Detailed Solution Below)
Sum Question 5 Detailed Solution
Formula used:
Sum of geometric progression (Sn) = {a × (rn - 1)}/(r - 1)
Where, a = first term ; r = common ratio ; n = number of term
Calculation:
3 +32 + 33 +...+ 38.
Here, a = 3 ; r = 3 ; n = 8
Sum of the series (S8) = {a × (r8 - 1)}/(r - 1)
⇒ {3 × (38 - 1)}/(3 - 1)
⇒ (3 × 6560)/2 = 3280 × 3
⇒ 9840
∴ The correct answer is 9840.
Sum Question 6:
The nth terms of the two series 3 + 10 + 17 + ... and 63 + 65 + 67 + .... are equal, then the value of n is:
Answer (Detailed Solution Below)
Sum Question 6 Detailed Solution
Given:
nth term for the first series = nth term for the second series.
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 and it can be positive, negative or zero.
Consider the AP, whose first term is 'a' and common difference is 'd' and there are n number of terms.
a, a + d, a + 2d, a + 3d........a + (n - 1)d
Hence, nth term of AP is given by
Tn = a + (n - 1)d
Calculation:
Consider the first series
3 + 10 + 17 + ...
a = 3 and d = 10 - 3 = 7
nth term of AP is given by,
(Tn)1 = 3 + (n - 1) × 7
⇒ (Tn)1 = 7n - 4 .......(1)
Consider the second series
63 + 65 + 67 + ....
a = 63 and d = 2
nth term of AP is given by,
(Tn)2 = 63 + (n - 1) × 2
⇒ (Tn)2 = 2n + 61 .......(2)
According to the question
(Tn)1 = (Tn)2
⇒ 7n - 4 = 2n + 61
⇒ 5n = 65
⇒ n = 13
Hence, for the given series, 13th term will be equal.
Additional Information
Sum of n terms of series, whose first term is a and the common difference is d.
\(S_n = \frac{n}{2}[ 2a + (n - 1)d]\)