QUERY 21

³⁄_1².2² + ⁵⁄_2².3² + ⁷⁄_3².4² + ⁹⁄_4².5² + ¹¹⁄_5².6² + ¹³⁄_6².7² + ¹⁵⁄_72.82 + ¹⁷⁄_8².9² + ¹⁹⁄_9².10² = ?

A) ¹⁄₁₀₀
B) ⁹⁹⁄₁₀₀
C) ¹⁰¹⁄₁₀₀
D) 1

MAHA GUPTA
³⁄_1².2² + ⁵⁄_2².3² + ⁷⁄_3².4² + ⁹⁄_4².5² + ¹¹⁄_5².6² + ¹³⁄_6².7² + ¹⁵⁄_7².8² + ¹⁷⁄_8².9² + ¹⁹⁄_9².10²

= ^2²-1²⁄_1².2² + ^3²-2²⁄_2².3² + ^4²-3²⁄_3².4² + ^5²-4²⁄_4².5² + … + ^10²-9²⁄_9².10²

= (1 – ¹⁄₄) + (¹⁄₄ – ¹⁄₉) + (¹⁄₉ – ¹⁄₁₆) + (¹⁄₁₆ – ¹⁄₂₅) + … + (¹⁄₈₁ – ¹⁄₁₀₀)

Cancelling similar terms, the above expression
= 1 – ¹⁄₁₀₀

= ⁹⁹⁄₁₀₀ (option ‘B’)

QUERY 22

The sum of 2n terms of the series: 1² – 2² + 3² – 4² + …

A) n²(2n + 1)
B) -n²(2n +1)
C) n(2n +1)
D) -n(2n +1)

MAHA GUPTA
Let’s see what is the 2n^th term of the above series. As every term is square of a number, so the 2n^th term will be (2n)²

Now we see that the set of every two terms of the series is in the form a² – b², so let’s include, in the above series, the term before (2n)². Let’s determine what it is.

We see in every pair i.e. (1² – 2²), (3² – 4²),  (5² – 6²), the first term is square of a number less by 1 of the square of the second term, so our required term is (2n – 1)². Now let’s write the series up to 2n^th term.

1² – 2² + 3² – 4² + … + (2n-1)² – (2n)²

= (1² – 2²) + (3² – 4²) + … + [(2n-1)² – (2n)²]

= (1 – 2)(1 + 2) + (3 – 4)(3 + 4) + … + [(2n-1) – 2n][(2n-1) + 2n]

= (-1)(1 + 2) + (-1)(3 + 4) + … + [(-1][(2n-1) + 2n]

= -1 – 2 – 3 – 4 – … – (2n-1) – 2n

= -[1 + 2 + 3 + 4 + … + (2n-1) + 2n]

We see that the series in the bracket is an A.P where a₁ (first term) = 1; a_n (last term) = 2n; t (number of terms) = 2n

Therefore,  -[1 + 2 + 3 + 4 + … + (2n-1) + 2n]
= ^t⁄₂(a₁ + a_n)
By putting value of these
= ²ⁿ⁄₂(1 + 2n)
= n(2n + 1) Option ‘C’

QUERY 23

If p, q, r are in Geometric Progression, then which is true among the following?

A) q = ^p+r⁄₂
B) p² = qr|
C) q = √(pr)
D) ^p⁄_r = ^r⁄_q

MAHA GUPTA
In a Geometric Progression, as ratio of any two consecutive terms is the same
Therefore,  _^p⁄q = ^q⁄_r
=> q² = pr

=> q = √(pr) (option ‘C’)

QUERY 24

Find the sum of the infinite series ¹⁄₂ + ¹⁄₄ + ¹⁄₈ + ¹⁄₁₆ + ¹⁄₃₂ …

A) 1
B) 2
C) ³⁄₄
D) ⁵⁄₄

MAHA GUPTA
The given series is a Geometric Progression with a (first term) = ¹⁄₂ and r (common ratio) =  _¹⁄2

We know that the sum of an infinite Geometric Progression = ^a⁄_{1 – r}

Therefore, the sum of the given series = ^1⁄2⁄_1-1⁄2

= 1 (option ‘A’)

QUERY 25

1+ ¹⁄₁₀ + ¹⁄_10² + ¹⁄_10³ + ¹⁄10⁴= ?

A) 1.1
B) 1.11
C) 1.111
D) 1.1111

MAHA GUPTA
1+ ¹⁄₁₀ + ¹⁄_10² + ¹⁄_10³ + ¹⁄₁₀⁴ 

= 1 + .1 + .001 + .0001 + .00001

= 1.1111 (option ‘D’)

NOTE: As this series is a G.P. you can solve it by using the G.P. formula, better to avoid that in this sum as it might be a little more time consuming.

QUERY 26

What is the sum of all the numbers between 200 and 600 which are divisible by 16?

A) 5000
B) 10000
C) 15000
D) 16000

MAHA GUPTA
200 = 16*12 + 8, means the number greater than 200 divisible by 16 = 16*13

600 = 3(16*12 +8) = 16*36 + 24 = 16*37 + 8, means the number smaller than 600 divisible by 16 = 16*37

Therefore, numbers between 200 and 600 which are divisible by 16
= (16*13) + (16*14) + (16*15) + … + (16*37)
= 16(13 + 14 + 15 + … + 37)

We see that the series in the bracket is an A.P where a₁ (first term) = 13; a_n (last term) = 37; t (number of terms) = 25

So the above expression
= 16[^t⁄₂(a₁ + a_n)]
= 16 [²⁵⁄₂(13 + 37)]
= 16*625
= 10000 (option ‘B’)

QUERY 27

Sum of all the three-digit numbers each of which on division by 5 leaves remainder 3 is?

A) 180
B) 1550
C) 6995
D) 99o9o

MAHA GUPTA
First term = 100+3 = 103
Last term = 995+3 = 998

So, the sum of the series
= 103 + 108 + 113 + 118 + … + 998

We see that this is a series in A.P where a₁ (first term) = 103; a_n (last term) = 998; t (number of terms) = ^998-103⁄₅ + 1 = 180

Therefore, 103 + 108 + 113 + 118 + … + 998 = ^t⁄₂(a₁ + a_n)
= ¹⁸⁰⁄₂(103 + 998)
= 99090 (option ‘D’)

 

QUERY 28

Raman earns Rs 10 on the first day, Rs 20 on the second day, Rs 40 on the third day, and so on. What will he earn in the first 6 days?

A) Rs 320
B) Rs 180
C) Rs 240
D) Rs 630

MAHA GUPTA
Total money earn by Raman = 10 + 20 + 40 + 80 + 160 + 320

We see that this is a G.P. where a (first term) = 10, r (common ration) = 2 , n (number of terms) = 6

Here r > 1, so sum of a G.P. where r > 1

= ^{a(rn – 1)}⁄_{r – 1}

= [10(2⁶-1)]/(2-1)

= 630 (option ‘D’)