General Series MCQ Quiz - Objective Question with Answer for General Series - Download Free PDF

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.

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.

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.

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.

Explanation:

Let no. of houses on each side = n

The sum of odd-numbered houses

= (301 + 303 + 305, …, n)

= \(\frac{n}{2}[2 \times 301+(n-1) 2] \)

= \(\frac{n}{2}[2 n+600]=n(n+300)\)

The sum of even-numbered houses

= 302 + 304 + 306, …, n

= \(\frac{n}{2}[2 \times 302+(n-1) 2] \)

= \(\frac{n}{2}[2 n+602]\)

= n[n + 301]

According to question

n[n + 301] – n[n + 300] = 27

n[n + 30] – n – 300] = 27

n = 27

n[n + 301] – n[n + 300] = 27

n[n + 30] – n – 300] = 27

n = 27

Alternate method:

If one house each side difference = 302 – 301 = 1

If two house each side difference = (302 + 304) – (301 + 303) = 2

If three houses each side difference = (302 + 304 + 306) – (301 + 303 + 39) = 3

So, to make difference equal to 27 number of houses on each side must be 27. 

Given:

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.

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.

Concept:

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.

Formula 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

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.

Given:

series = 7, 11, 15, 19, 23, …

Formula Used:

S_n = n / 2 × {2a + (n - 1) × d}; where n= number of terms, d = common difference, a = first term, S_n = 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.

Given:

n = 9, d = 2 and Sum = 558

Formula used:

Sum of n term of AP = (n/2)(2a + (n - 1)d)

n^th 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

Explanation:

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.

Formula used:

n^th 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

n^th 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

Given:

1 + 4 + 9 + 16 + 25 + ..... + 81 + 100 + 81 + ...... + 16 + 9 + 4 + 1 

Concept used:

Sum of Square of 1^st  '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 1^st  '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.