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SUM OF SERIES, SUM OF NUMBERS AND SUM OF DIGITS

QUERY 11

Find the value of 1 × 2 + 2 × 3 + 3 × 4 + 4 × 5 + 5 × 6 + 6 × 7 + 7 × 8 + 8 × 9 + 9 × 10 + 10 × 11 

A) 770
B) 660
C) 440
D) 330

MAHA GUPTA
1 × 2 + 2 × 3 + 3 × 4 + 4 × 5 + 5 × 6 + 6 × 7 + 7 × 8 + 8 × 9 + 9 × 10 + 10 × 11
= (1 × 2) + (2 × 3) + (3 × 4) + (4 × 5) + (5 × 6) + (6 × 7) + (7 × 8) + (8 × 9) + (9 × 10) + (10 × 11)

Sum of such kind of a series = ^n(n+1)(n+2)⁄₃; where n is the number of terms

We see n = 10 here

Therefore sum of the series = ^{10(10+1)(10+2)}⁄₃
= ^{10 × 11 × 12}⁄₃

= 440 (option ‘C’)

QUERY 11

The sum of three digit numbers is? 

A) 98901
B) 494550
C) 8991
D) 899

MAHA GUPTA
Sum of three digit numbers = 100 + 102 + 103 + 104 + … + 999
Here number of terms (n) = 999 – 99 = 900
First term (a) = 101
Last term (l) = 999

Now, the sum = ^{n(a + l)}⁄₂

= ^{900(100 + 999)}⁄₂
= 494550 (option ‘B’)

QUERY 12

Find: 2² + 3² + 4² + … + 10²

A) 385
B) 2916
C) 540
D) 384

MAHA GUPTA
2² + 3² + 4² + … + 10²
= (1² + 2² + 3² + 4² + … + 10²) – 1²

Sum of a series of this type 1² + 2² + 3² + 4² + … + 10² = ^n(n+1)(2n+1)⁄₆

Therefore, the required sum = ^{10(10+1)(2*10+1)}⁄₆ – 1²

= ^{(10 ×11 × 21)}⁄₆ – 1

= 384 (option ‘D’)

QUERY 13

If 1³ + 2³ + 3³ + … + 9³ = 2025, then 0.11³ + 0.22³ + … + 0.99³ = ?

A) 0.2695
B) 2.695
C) 3.695
D) 0.3695

MAHA GUPTA
0.11³ + 0.22³ + … + 0.99³
= 0.11³(1³ + 2³ + 3³ + … + 9³)
= 0.11³ × 2025
= 2.695275
= 2.695 (option ‘B’)

QUERY 14

In a book, containing 20 pages, one sheet is missing. Sum of the page numbers of the remaining pages is 195. The numbers written on both the sides of the missing sheet must be?

A) 5, 6
B) 6, 7
C) 7, 8
D) 8, 9

MAHA GUPTA
Sum of all pages including missing pages = Sum of natural numbers from 1 to 20
= n(n+1)/2 = 20 (20+1)/2 = 210

Sum of page numbers excluding missing numbers = 195 (given)

Therefore, sum of page numbers of the missing sheet = 210 – 195 = 15

But page numbers of a sheet are always consecutive
Hence, numbers written on both sides of the missing sheet = 7 and 8 (option ‘C’)

QUERY 15

Simplify this: 999¹⁄₇ + 999²⁄₇ + 999³⁄₇ + 999⁴⁄₇ + 999⁵⁄₇ + 999⁶⁄₇

A) 5997
B) 5979
C) 5994
D) 2997

MAHA GUPTA
999¹⁄₇ + 999²⁄₇ + 999³⁄₇ + 999⁴⁄₇ + 999⁵⁄₇ + 999⁶⁄₇

= 999+¹⁄₇ + 999+²⁄₇ + 999+³⁄₇ + 999+⁴⁄₇ + 999+⁵⁄₇ + 999+⁶⁄₇

As 999 has been added 6 times, so the above expression
= 999*6 + (¹⁄₇ + ²⁄₇ + ³⁄₇ + ⁴⁄₇ + ⁵⁄₇ + ⁶⁄₇)

But, ¹⁄₇ + ²⁄₇ + ³⁄₇ + ⁴⁄₇ + ⁵⁄₇ + ⁶⁄₇ is an A.P. in which a (first term) = ¹⁄₇, l (last term)= ⁶⁄₇, n (number of terms) = 6

Therefore, ¹⁄₇ + ²⁄₇ + ³⁄₇ + ⁴⁄₇ + ⁵⁄₇ + ⁶⁄₇ = ⁿ⁄₂(a + l) = ⁶⁄₂(¹⁄₇ + ⁶⁄₇) = 3

Hence, 999*6 + (¹⁄₇ + ²⁄₇ + ³⁄₇ + ⁴⁄₇ + ⁵⁄₇ + ⁶⁄₇) = 5594 + 3 = 5997 (option ‘A’)

QUERY 16

Simplify this: (1 – ¹⁄₃)(1 – ¹⁄₄)(1 – ¹⁄₅) … (1 – ¹⁄_n)

A) ¹⁄_n
B) ²⁄_n
C) ^2(n-1)⁄_n
D) ²⁄_n(n+1)

MAHA GUPTA
(1 – ¹⁄₃)(1 – ¹⁄₄)(1 – ¹⁄₅) … (1 – ¹⁄_n)
= ²⁄₃ × ³⁄₄ × ⁴⁄₅ × … × ^n-1⁄_n

= By putting any value of n, you’ll find that the denominator if the first number will cancel by the numerator of the next number, and so on. In the end we will be left only with the numerator of the first number and denominator of the last number.

Hence, ²⁄₃ × ³⁄₄ × ⁴⁄₅ × … × ^n-1⁄_n = 2/n (option ‘B’)

QUERY 17

(1 + ¹⁄_x+1)(1 + ¹⁄_x+2)(1 + ¹⁄_x+3)(1+ ¹⁄_x+4) = ?

A) 4 + ¹⁄_x+1
B) 4 + ¹⁄_x+4
C) ^x+5⁄_x+1
D) ^x+2⁄_x+4

MAHA GUPTA
(1 + ¹⁄_x+1)(1 + ¹⁄_x+2)(1 + ¹⁄_x+3)(1+ ¹⁄_x+4)

= (^x+2⁄_x+1) × ( ^x+3⁄_x+2 ) × (^x+4⁄_x+3 ) × (^x+5⁄_x+4)

Cancelling similar numerators and denominators, the above expression
= ^x+5⁄_x+1 (option ‘C’)

QUERY 18

¹⁄₂₀ + ¹⁄₃₀ + ¹⁄₄₂ + ¹⁄₅₆ + ¹⁄₇₂ + ¹⁄₉₀ + ¹⁄₁₁₀ + ¹⁄₁₃₂ is equal to?

A) ¹⁄₈
B) ¹⁄₇
C) ¹⁄₆
D) ¹⁄₁₀

MAHA GUPTA
¹⁄₂₀ + ¹⁄₃₀ + ¹⁄₄₂ + ¹⁄₅₆ + ¹⁄₇₂ + ¹⁄₉₀ + ¹⁄₁₁₀ + ¹⁄₁₃₂

= ¹⁄_4×5 + ¹⁄_5×6 + ¹⁄_6×7 + … + ¹⁄_11×12

= (¹⁄₄ – ¹⁄₅) + (¹⁄₅ – ¹⁄₆) + (¹⁄₆ – ¹⁄₇) + … + (¹⁄₁₁ – ¹⁄₁₂)

= ¹⁄₄ – ¹⁄₅ + ¹⁄₅ – ¹⁄₆ + ¹⁄₆ – ¹⁄₇ + … + ¹⁄₁₁ – ¹⁄₁₂

Cancelling similar terms with plus and minus signs the above expression

= ¹⁄₄ – ¹⁄₁₂ = ¹⁄₆ (option ‘C’)

QUERY 19

¹⁄_1×4 + ¹⁄_4×7 + ¹⁄_7×10 + ¹⁄_10×13 + ¹⁄_13×16 = ?

A) ¹⁄₃
B) ⁵⁄₁₆
C) ³⁄₈
D) ⁴¹⁄₇₂₈₀

MAHA GUPTA
To solve such types of series with ease, make sure that the numerator and the difference of the denominator of each fraction is similar. For example, here in the above series the numerator of the each fraction is 1 whereas the difference of  the denominators of each fraction is 3, means they are not similar. For that we need to multiply the numerator of each fraction by 3. Obviously in that case, we’ll have to multiply the whole series by ¹⁄₃. Now let’s start solving the series.

¹⁄_1×4 + ¹⁄_4×7 + ¹⁄_7×10 + ¹⁄_10×13 + ¹⁄_13×16

= ¹⁄₃(³⁄_1×4 + ³⁄_4×7 + ³⁄_7×10 + ³⁄_10×13 + ³⁄_13×16)

= ¹⁄₃(1 – ¹⁄₄) + (¹⁄₄ – ¹⁄₇) + (¹⁄₇ – ¹⁄₁₀) + (¹⁄₁₀ – ¹⁄₁₃) + (¹⁄₁₃ – ¹⁄₁₆)

By cancelling the similar terms the above series
= ¹⁄₃(1 – ¹⁄₁₆)

= ⁵⁄₁₆ (option ‘B’)

QUERY 20

(1 – ¹⁄_2²)(1 – ¹⁄_3²) … (1 – ¹⁄_9²)(1 – ¹⁄_10²) = ?

A) ¹¹⁄₂₀
B) ¹⁄₂₀
C) ¹⁄₂
D) ¹⁰⁄₂₁

MAHA GUPTA
(1 – ¹⁄_2²)(1 – ¹⁄_3²) … (1 – ¹⁄_9²)(1 – ¹⁄_10²) = ?

Applying a² – b² = (a – b)(a + b), as each expression in brackets of the above series is in the form a² – b²
= (1 – ¹⁄₂)(1 + ¹⁄₂)(1 – ¹⁄₃)(1 + ¹⁄₃) … (1 – ¹⁄₉)(1 + ¹⁄₉)(1 – ¹⁄₁₀)(1 + ¹⁄₁₀)

= (1 – ¹⁄₂)(1 – ¹⁄₃) … (1 – ¹⁄₉)(1 – ¹⁄₁₀) × (1 + ¹⁄₂)(1 + ¹⁄₃) … (1 + ¹⁄₉)(1 + ¹⁄₁₀)

= (¹⁄₂ × ²⁄₃ × … × ⁸⁄₉ × ⁹⁄₁₀) × (³⁄₂ × ⁴⁄₃ × … × ¹⁰⁄₉ × ¹¹⁄₁₀)

Cancelling similar denominators and numerators from the above the expression
= ¹⁄₁₀ × ¹¹⁄₂

= ¹¹⁄₂₀ (option ‘A’)