General Series MCQ Quiz - Objective Question with Answer for General Series - Download Free PDF
Last updated on Jun 11, 2025
Latest General Series MCQ Objective Questions
General Series Question 1:
If N1 = 3 + 33 + 333 + .... + 333333 and N2 = 4 + 44 + 444 + .... + 4444444, then what is the sum of digits of 'N1 + N2'?
Answer (Detailed Solution Below)
General Series Question 1 Detailed Solution
Given:
N1 = 3 + 33 + 333 + .... + 333333 and
N2 = 4 + 44 + 444 + .... + 4444444
Calculations:
N1 = 3 + 33 + 333 + 3333 + 33333 + 333333
⇒ N1 = 3 × (1 + 11 + 111 + 1111 + 11111 + 111111)
Sum of series: 1 + 11 + 111 + 1111 + 11111 + 111111 = 123456
⇒ N1 = 3 × 123456 = 370368
N2 = 4 + 44 + 444 + 4444 + 44444 + 444444 + 4444444
⇒ N2 = 4 × (1 + 11 + 111 + 1111 + 11111 + 111111 + 1111111)
Sum of series: 1 + 11 + 111 + 1111 + 11111 + 111111 + 1111111 = 1234567
⇒ N2 = 4 × 1234567 = 4938268
N1 + N2 = 370368 + 4938268
⇒ N1 + N2 = 5308636
Sum of digits of 5308636 ⇒ 5 + 3 + 0 + 8 + 6 + 3 + 6 = 31
The correct answer is option 2.
General Series Question 2:
Simplify the following.
\({\left[ {\left( {1 - \frac{1}{2}} \right)\left( {1 - \frac{1}{3}} \right)\left( {1 - \frac{1}{4}} \right) \ldots \left( {1 - \frac{1}{{100}}} \right)} \right]^{ - 0.5}}\)
Answer (Detailed Solution Below)
General Series Question 2 Detailed Solution
Given:
\({\left[ {\left( {1 - \frac{1}{2}} \right)\left( {1 - \frac{1}{3}} \right)\left( {1 - \frac{1}{4}} \right) \ldots \left( {1 - \frac{1}{{100}}} \right)} \right]^{ - 0.5}}\)
Calculations:
\({\left[ {\left( {1 - \frac{1}{2}} \right)\left( {1 - \frac{1}{3}} \right)\left( {1 - \frac{1}{4}} \right) \ldots \left( {1 - \frac{1}{{100}}} \right)} \right]^ { - 0.5}}\)
⇒ \([{(\frac{1}{2})(\frac{2}{3}) (\frac{3}{4}).....(\frac{99}{100})]^{-0.5} }\)
After solving, we get
⇒ (1/100)-0.5
⇒ (100)0.5
⇒ 10
∴ The value of \({\left[ {\left( {1 - \frac{1}{2}} \right)\left( {1 - \frac{1}{3}} \right)\left( {1 - \frac{1}{4}} \right) \ldots \left( {1 - \frac{1}{{100}}} \right)} \right]^{ - 0.5}}\) is 10.
General Series Question 3:
\(\text{If } T_n = T_{n-1} + T_{n-2} \quad \forall\, n \geq 3 \text{ and } T_1 = T_2 = 1 \text{ then } T_1 + T_2 + \ldots + T_{10} =\ ? \)
Answer (Detailed Solution Below)
General Series Question 3 Detailed Solution
Given:
If Tn = Tn-1 + Tn-2 ∀ n ≥ 3 and T1 = T2 = 1, then find T1 + T2 + ... + T10.
Formula used:
Fibonacci series formula: Tn = Tn-1 + Tn-2
Sum of terms: S = T1 + T2 + ... + T10
Calculation:
Calculate Fibonacci terms up to T10
T1 = 1
T2 = 1
T3 = T2 + T1 = 1 + 1 = 2
T4 = T3 + T2 = 2 + 1 = 3
T5 = T4 + T3 = 3 + 2 = 5
T6 = T5 + T4 = 5 + 3 = 8
T7 = T6 + T5 = 8 + 5 = 13
T8 = T7 + T6 = 13 + 8 = 21
T9 = T8 + T7 = 21 + 13 = 34
T10 = T9 + T8 = 34 + 21 = 55
Calculate the sum S
⇒ S = T1 + T2 + T3 + T4 + T5 + T6 + T7 + T8 + T9 + T10
⇒ S = 1 + 1 + 2 + 3 + 5 + 8 + 13 + 21 + 34 + 55 = 143
Option 2: T12 - 1
⇒ T12 = T11 + T10 = 89 + 55 = 144
⇒ T12 - 1 = 144 - 1 = 143
The Correct answer is Option 2.
General Series Question 4:
If \(\rm S=\frac{3}{2}+\frac{5}{4}+\frac{7}{8}+...+\frac{2n+1}{2^n}\), where n = 1, 2, 3...then the value of S is
Answer (Detailed Solution Below)
General Series Question 4 Detailed Solution
Given:
\(\rm S=\frac{3}{2}+\frac{5}{4}+\frac{7}{8}+...+\frac{2n+1}{2^n}\)
Calculation:
\(\rm S = \frac{3}{2} + \frac{5}{4} + \frac{7}{8} + \frac{9}{16} + ... + \frac{2n+1}{2^n}\)
Multiply by \(\rm \frac{1}{2}\):
\(\rm \frac{1}{2}S = \frac{3}{4} + \frac{5}{8} + \frac{7}{16} + ... + \frac{2n-1}{2^n} + \frac{2n+1}{2^{n+1}}\)
Subtract the second equation from the first:
\(\rm S - \frac{1}{2}S = \frac{3}{2} + \left(\frac{5}{4} - \frac{3}{4}\right) + \left(\frac{7}{8} - \frac{5}{8}\right) + \left(\frac{9}{16} - \frac{7}{16}\right) + ... + \left(\frac{2n+1}{2^n} - \frac{2n-1}{2^n}\right) - \frac{2n+1}{2^{n+1}}\)
\(\rm \frac{1}{2}S = \frac{3}{2} + \frac{2}{4} + \frac{2}{8} + \frac{2}{16} + ... + \frac{2}{2^n} - \frac{2n+1}{2^{n+1}}\)
\(\rm \frac{1}{2}S = \frac{3}{2} + \left(\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + ... + \frac{1}{2^{n-1}}\right) - \frac{2n+1}{2^{n+1}}\)
The terms in the parentheses form a geometric series with first term \(\rm a = \frac{1}{2}\), common ratio \(\rm r = \frac{1}{2}\), and \(\rm n-1\) terms.
Sum of the geometric series = \(\rm \frac{a(1 - r^{n-1})}{1 - r} = \frac{\frac{1}{2}\left(1 - \left(\frac{1}{2}\right)^{n-1}\right)}{1 - \frac{1}{2}} = \frac{\frac{1}{2}\left(1 - \frac{1}{2^{n-1}}\right)}{\frac{1}{2}} = 1 - \frac{1}{2^{n-1}}\)
\(\rm \frac{1}{2}S = \frac{3}{2} + \left(1 - \frac{1}{2^{n-1}}\right) - \frac{2n+1}{2^{n+1}}\)
\(\rm \frac{1}{2}S = \frac{3}{2} + 1 - \frac{2}{2^n} - \frac{2n+1}{2^{n+1}}\)
\(\rm \frac{1}{2}S = \frac{5}{2} - \frac{4}{2^{n+1}} - \frac{2n+1}{2^{n+1}}\)
\(\rm \frac{1}{2}S = \frac{5}{2} - \frac{4 + 2n + 1}{2^{n+1}}\)
\(\rm \frac{1}{2}S = \frac{5}{2} - \frac{2n + 5}{2^{n+1}}\)
\(\rm S = 5 - \frac{2n + 5}{2^{n}}\)
For the infinite sum as \(\rm n \to \infty\), \(\rm \frac{2n + 5}{2^n} \to 0\).
Therefore, \(\rm S = 5 - 0 = 5\) (for the infinite series).
∴ The value of S is 5.
General Series Question 5:
Suppose the nth term of a series is \(1+\frac{n}{2}+\frac{n^{2}}{2} \). If there are 20 terms in the series, then the sum of the series is equal to
Answer (Detailed Solution Below)
General Series Question 5 Detailed Solution
Given:
nth term = 1 + n/2 + n2/2
Sum of natural numbers = n(n+1)/2
Sum of square of 'n' natural numbers = [n(n+1)(2n+1)]/6
Calculation:
For understanding, breaking the terms given in question as :
1. Adding one twenty times = 20
2. When n = 1, 2, 3 and so on up to 20 in n/2, the sum = 1/2 (n(n+1)/2)
= 1/2 (20(20+1)/2) = 105
3. When n = 1, 2, 3 and so on up to 20 in n2/2, the sum = 1/2 [n(n+1)(2n+1)]/6
= 1/2 [20(20+1)(2×20+1)]/6 = 1435
Therefore, sum of the series = 20 + 105 + 1435 = 1560
Hence, option 4 is correct.
Top General Series MCQ Objective Questions
What is the sum of first 20 terms of the following series ?
1 × 2 + 2 × 3 + 3 × 4 + 4 × 5 + ..........
Answer (Detailed Solution Below)
General Series Question 6 Detailed Solution
Download Solution PDFGiven:
1 × 2 + 2 × 3 + 3 × 4 + 4 × 5 + .......... + 20
Concept used:
The sum of the first n consecutive no. is n(n+1)/2.
The sum of the square of the first n consecutive no. is n(n + 1)(2n + 1)/6.
Calculations:
According to question, we have:
1 × 2 + 2 × 3 + 3 × 4 + 4 × 5 + ..........up to 20 terms
nth term = n(n + 1)
Sum = ∑ n (n + 1) = ∑ (n2 + n)
Sum = ∑ n2 + ∑ n
Sum = \(\frac{n(n +1)(2n +1)}{6}+\frac{n(n +1)}{2}\)
Sum = \(\frac{n(n +1)}{2}\Big[\frac{2n +1}{3}+1\Big]\)
Sum = \(\frac{n(n +1)}{2}\Big[\frac{2n +4}{3}\Big]\)= \(\frac{n(n +1)(2n + 4)}{6}\)
Substitute n = 20, we have
Sum = \(\frac{20(20 +1)(2\times 20+ 4)}{6}\)
⇒ Sum = \(\frac{20\times 21 \times 44}{6}\)
⇒ Sum = \(\frac{420 \times 44}{6}\)
⇒ Sum = 3080
∴ The sum of first 20 terms of the given series is 3080.
The sum of 1 - 1 + 1 - 1 + 1 - 1... to even number of terms is ____.
Answer (Detailed Solution Below)
General Series Question 7 Detailed Solution
Download Solution PDFConcept:
Even: Any number in which digits at once place is 0, 2, 4, 6 or 8.
For example 2, 48, 1000 are even number.
Calculation:
Given number series is
1 - 1 + 1 - 1 + 1 - 1......
We can see that numbers changes sign at every position and, the sign of numbers in alternate place are same. Hence,
Sum of first two terms
= 1 - 1 = 0
Sum of next two terms
= 1 - 1 = 0
We can see that sum of every pair is zero.
∴ For an even number of total terms, the value of the series will be zero.
Suppose the nth term of a series is \(1+\frac{n}{2}+\frac{n^{2}}{2} \). If there are 20 terms in the series, then the sum of the series is equal to
Answer (Detailed Solution Below)
General Series Question 8 Detailed Solution
Download Solution PDFGiven:
nth term = 1 + n/2 + n2/2
Sum of natural numbers = n(n+1)/2
Sum of square of 'n' natural numbers = [n(n+1)(2n+1)]/6
Calculation:
For understanding, breaking the terms given in question as :
1. Adding one twenty times = 20
2. When n = 1, 2, 3 and so on up to 20 in n/2, the sum = 1/2 (n(n+1)/2)
= 1/2 (20(20+1)/2) = 105
3. When n = 1, 2, 3 and so on up to 20 in n2/2, the sum = 1/2 [n(n+1)(2n+1)]/6
= 1/2 [20(20+1)(2×20+1)]/6 = 1435
Therefore, sum of the series = 20 + 105 + 1435 = 1560
Hence, option 4 is correct.
The sum of all even numbers between 1 and 55 is
Answer (Detailed Solution Below)
General Series Question 9 Detailed Solution
Download Solution PDFFormula used:
Sum of n terms of an AP = (n/2) [a + l] = (n/2) [2a + (n - 1)d]
Where n = number of terms, d = Common difference, l = Last term, and a = first term
Calculation:
Even numbers between 1 and 55
a = 2, l = 54
Sum of even numbers = (n/2) [a + l]
Here, n is number of even terms = 27
⇒ (27/2) [2 + 54]
⇒ 27 × 28
⇒ 756
∴ The sum of all even numbers between 1 and 55 is 756
Simplify the following.
\({\left[ {\left( {1 - \frac{1}{2}} \right)\left( {1 - \frac{1}{3}} \right)\left( {1 - \frac{1}{4}} \right) \ldots \left( {1 - \frac{1}{{100}}} \right)} \right]^{ - 0.5}}\)
Answer (Detailed Solution Below)
General Series Question 10 Detailed Solution
Download Solution PDFGiven:
\({\left[ {\left( {1 - \frac{1}{2}} \right)\left( {1 - \frac{1}{3}} \right)\left( {1 - \frac{1}{4}} \right) \ldots \left( {1 - \frac{1}{{100}}} \right)} \right]^{ - 0.5}}\)
Calculations:
\({\left[ {\left( {1 - \frac{1}{2}} \right)\left( {1 - \frac{1}{3}} \right)\left( {1 - \frac{1}{4}} \right) \ldots \left( {1 - \frac{1}{{100}}} \right)} \right]^ { - 0.5}}\)
⇒ \([{(\frac{1}{2})(\frac{2}{3}) (\frac{3}{4}).....(\frac{99}{100})]^{-0.5} }\)
After solving, we get
⇒ (1/100)-0.5
⇒ (100)0.5
⇒ 10
∴ The value of \({\left[ {\left( {1 - \frac{1}{2}} \right)\left( {1 - \frac{1}{3}} \right)\left( {1 - \frac{1}{4}} \right) \ldots \left( {1 - \frac{1}{{100}}} \right)} \right]^{ - 0.5}}\) is 10.
Sum of first 25 terms in the following series would be
7, 11, 15, 19, 23, …Answer (Detailed Solution Below)
General Series Question 11 Detailed Solution
Download Solution PDFGiven:
series = 7, 11, 15, 19, 23, …
Formula Used:
Sn = n / 2 × {2a + (n - 1) × d}; where n= number of terms, d = common difference, a = first term, Sn = sum of n terms; d = (a2 - a1); a2 = second term of A.P., a1 = first term of an A.P
Calculation:
Using the above formulae, we get d = 4
⇒ S25 = 25 / 2 × {2 × 7 + (25 - 1) × 4}
⇒ S25 = 55 × 25
⇒ S25 = 1375
Required sum of 25 terms = 1375
Therefore the correct answer is 1375.
The sum of 9 even consecutive number 558. What is the sum of the last 3 consecutive numbers?
Answer (Detailed Solution Below)
General Series Question 12 Detailed Solution
Download Solution PDFGiven:
n = 9, d = 2 and Sum = 558
Formula used:
Sum of n term of AP = (n/2)(2a + (n - 1)d)
nth term = a + (n - 1)d
Where, a = first term, d = common difference, n = total number of terms
Calculation:
Sum of n term of AP = (n/2)(2a + (n - 1)d)
⇒ 558 = (9/2)(2a + (9 - 1) × 2)
⇒ 558 = (9/2)(2a + 16)
⇒ 124 = 2a + 16
⇒ 2a = 108
⇒ a = 54
∴ The 9 consecutive even numbers are 54, 56, 58, 60, 62, 64, 66, 68, 70
Sum of last 3 consecutive number = 66 + 68 + 70 = 204
if arithmetic mean, geometric mean and harmonic mean between two numbers a and b are A, G & H then A, G H will be
Answer (Detailed Solution Below)
General Series Question 13 Detailed Solution
Download Solution PDFExplanation:
If A is the arithmetic mean between a and b,
\(⇒ A=\frac{a+b}{2}\)
If G is the geometric mean between a and b,
⇒ G = √ab
If H is the harmonic mean between a and b,
\(⇒ H=\frac{2ab}{a+b}\)
Now, AH = \(\frac{a+b}{2}\times \frac{2ab}{a+b}\)
⇒ AH = ab
⇒ AH = G2
This is a form of a G.P.
Hence, If the arithmetic mean, geometric mean, and Harmonic mean between two numbers 'a' and 'b' are A, G, and H respectively, then A, G, H will be in geometric series.
The value of 65 + 67 + 69 + ... + 135 = ?
Answer (Detailed Solution Below)
General Series Question 14 Detailed Solution
Download Solution PDFFormula used:
nth term = a + (n - 1)d
Sum = \({n\over2}({a + l})\)
Here, a → First term, n → Total number of terms, d → Common difference, l → Last term
Calculation:
Our given series is 65 + 67 + 69 + ... + 135
First term, (a) = 65,
Common differnce = 67 - 65 = 2
nth term = 135
Now, nth term = a + (n - 1)d
⇒ 135 = 65 + (n - 1) × 2
⇒ 2(n - 1) = 135 - 65
⇒ 2(n - 1) = 70
⇒ n = 36
Sum = \({n\over2}({a + l})\)
⇒ (36/2)(65 + 135)
⇒ 3600
1 + 4 + 9 + 16 + 25 + ..... + 81 + 100 + 81 + ...... + 16 + 9 + 4 + 1 = ?
Answer (Detailed Solution Below)
General Series Question 15 Detailed Solution
Download Solution PDFGiven:
1 + 4 + 9 + 16 + 25 + ..... + 81 + 100 + 81 + ...... + 16 + 9 + 4 + 1
Concept used:
Sum of Square of 1st 'n' natural numbers is given by
12 + 22 + 32 + ..... + n2 = \(\frac{n(n+1)(2n+1)}{6}\)
Calculation:
According to the question,
1 + 4 + 9 + 16 + 25 + ..... + 81 + 100 + 81 + ..... + 16 + 9 + 4 + 1
⇒ 2 (1 + 4 + 9 + 16 + 25 + ..... + 81) + 100
⇒ 2 (12 + 22 + 32 + 42 + .... + 92) + 100
By using the formula for sum of squares of 1st '9' natural numbers,
⇒ 2 \( [\frac{9(9+1)(18 +1)}{6}]\) + 100
⇒ 2 \((\frac{9\; \times \;10 \;\times \;19}{6} )\) + 100
⇒ 570 + 100 = 670
∴ The sum of the given series is 670.