# SUM OF SERIES, SUM OF NUMBERS AND SUM OF DIGITS (PART-I)

Click Here to Buy Online Practice Sets exclusivly on Testbook.com

## SUM OF SERIES, SUM OF NUMBERS AND SUM OF DIGITS

#### QUERY 1

**Find the sum of all the digits from 1 to 100 is.**

A) 109

B) 900

C) 901

D) 902

**MAHA GUPTA
**Sum of digits of the numbers from 1 to 99

= 10(1 + 2 + 3 +……….+ 9) + 10(1 + 2 + 3 +……….+ 9)

= 10 x 45 + 10 x 45

= 450 + 450 = 900

But we have to add the digit of the numbers from 1 to 100

Now, sum of the digits in 100 = 1+0+0 = 1

Therefore, sum of the digits of the numbers from 1 to 100

900+1 = 901 (option ‘C’)

#### QUERY 2

**Three numbers non-divisible by each other are such that the product of the first two is 551 and product of the last two is 1073. Find their sum.**

A) 75

B) 81

C) 85

D) 89

**Vaibhav Vats**

Find the factors of the product of first two numbers and as well as last two numbers, you’ll then find a common factor in those, obviously it will be the number in the middle of those three numbers

Now the factors of the product 551 = 19*29

and the factors of the product 1073 = 29 *37

We see that the common factor is 29, so the numbers are 19, 29, and 37

Therefore the sum = 19 + 29 + 37 = 85 (option ‘C’)

#### QUERY 3

**If we write the numbers from 50 to 151 what is the difference between sum of all odd and even numbers?**

A) 48

B) 49

C) 50

D) 51

**Amit jha**

In such a question the total of the numbers will always be in even, which could be found in one of the manners below:

The greatest number – (The smallest number – 1)

So the total of the numbers in the given question = 151 – (50 – 1) = 151 – 49 = 102

Obviously half of them will be odd and half even

So 102/2 i.e. 51 of them are odd and 51 are even

Now find the difference of each pair so got

You’ll see it’s 1 in every case

But we have 51 such pairs in total

So the difference between the sum of all all odd and even numbers from 50 to 151 = 51 (option ‘D’)

#### QUERY 4

**Find the sum of first 50 terms in the series: 5 + 3 – 7 + 5 + 3 …**

A) 1

B) 8

C) 24

D) 40

**MAHA GUPTA
**The first three terms i.e. (5 + 3 – 7) are making one set of the series

Let’s see how many such sets are there in the series

Number of terms given = 50

So, the number of sets = 50/3 = 16 +2, means 16 is the number of full sets of three terms each and 2 being the first two terms.

Sum of one set = 5 + 3 – 7 = 1

Sum of the next 2 terms = 5 + 3 = 8

Hence the sum of all 50 terms = Sum of all full sets + Sum of the remaining 2 terms

= 16*1 + 8 = 24 (option ‘C’)

#### QUERY 5

**The number 25 ^{64} **

**x**

**64**

^{25}is square of a natural number n. Find the sum of the digits of the number n.A) 10

B) 14

B) 11

B) 25

**Sumer Singh Chouhann**

n² = 25^{64} x 64^{25
}=> n² = (5²)^{64} x (2^{6})^{25}

=> n^{2} = (5^{64})^{2} x 2^{150}

=> n^{2} = (5^{64})^{2} x 2^{(128+22)}

=> n^{2} = (5^{64})^{2} x (2^{64})^{2} x (2^{11})^{2}

=> n = 5^{64} x 2^{64} x 2^{11}

=> n =10^{64} x 2048

If 10 raised to power any natural number multiplied with a number gives only zeros at the right, so the sum of the digit of the number n = 2 + 0 + 4+ 8 = 14 (option ‘B’)

#### QUERY 6

**61 + 62 + 63 + … + 100 = ?**

A) 2525

B) 2975

C) 3220

D) 3775

**MAHA GUPTA**

This is the series of addition of some consecutive natural numbers. In such a series if the numbers to be added is even, the sum of the series is found out by multiplying the number of pairs by addition of the greatest number and the small number.

Now number of pairs = (1/2)/[(100-61) + 1]

=> (1/2)/40

=> 20

Therefore, sum of the series = 20 x (100+61)

= 20 x 161

= 3220 (option ‘C’)

Another Method

61 + 62 + 63 + … + 100

In the above series n = 40, the first term (a) = 61 and the last term (l) = 100

Therefore the sum = ^{n(a + l)}⁄_{2}

= ^{40(60 + 100)}⁄_{2}

= 3220 (option ‘C’)

One more method

61 + 62 + 63 + … + 100

= (1 + 2 + 3 + … + 100) – (1 + 2 + 3 + … + 60)

= ^{100}⁄_{2}(100 + 1) – ^{60}⁄_{2}(60 + 1) [using n = ^{n}⁄_{2}(n+1)]

= (50 x 101) – (30 x 61)

= (5050 – 183o)

= 3220 (option ‘C’)

#### QUERY 7

**2² + 4² + 6² + … + 20² = ?**

A) 770

B) 1155

C) 1540

D) 385 × 385

**MAHA GUPTA
**2² + 4² + 6² + … + 20² = 2²(1² + 2² + 3² + … + 10²)

= 4(

^{1}⁄

_{6}× 10 × 11 × 21) [As 1² + 2² + 3² + … + n² =

^{1}⁄

_{6}n(n + 1)(2n + 1)]

= 1540.

#### QUERY 8

**How many terms are there in 2, 4, 8, 16, …, 1024?**

A) 8

B) 10

C) 15

D) 80

**MAHA GUPTA**

You see that this a G.P with a = 2 and r = 4/2 = 2; where **a** is the first term and **r** the common ratio.

The **n-th term** of a geometric sequence with the first term **a** and common ratio **r** is given by

A_{n} = ar^{n-1}

Therefore, 1024 = 2*2^{n-1}

=> 2^{n-1 }= 512

=> 2^{n-1 }= 2^{9}

=> n – 1 = 9

=> n = 10 (option ‘B’)

**TRICK**

In this question you see that every term of the series is ‘2 raised to the power of a consecutive natural number beginning with 1’, so our answer will be 10 as 2^{10} = 1024

#### QUERY 9

**2 + 2 ^{2} + 2^{3} + … + 2^{8}**

A) 510

B) 1020

C) 1530

D) 400

**MAHA GUPTA**

You see that this a G.P with a = 2 and r = 2²⁄_{2} = 2; r^{n} = 2^{8 }where **a** is the first term, **r** the common ratio and **n** is the number of terms.

**Sum of n terms in a G.P.**

**S _{n} = ^{a(rn – 1)}⁄_{r – 1}, if r > 1**

**S _{n }= ^{a(1 – rn)}⁄_{1 – r}, if r < 1**

**Sum of infinite terms in a G.P.**

**S _{∞ }= ^{a}⁄_{1 – r}, if (| r | < 1 )**

Here as r > 1, so sum of the given series S_{n} = ^{a(rn-1)}⁄_{r–1}

= ^{2*(28 – 1)}⁄_{2-1}

= ^{2*(256 – 1)}⁄_{1}

= 2*255

= 510 (option ‘A’)

#### QUERY 10

**Find the sum of the series
**

**1 +**

^{4}**⁄**

_{7}**+**

^{9}**⁄**

_{7²}+

^{16}**⁄**

_{7³}+ ………infiniteA) ^{49}⁄_{27}

B) ^{36}⁄_{49}

C) ^{49}⁄_{36}

D) ^{27}⁄_{49}

**MAHA GUPTA
**It can easily be observed that this series can be converted into a geometric series, so our focus will be on this point while solving it:

Now let x = 1 +

^{4}⁄

_{7}+

^{9}⁄

_{7²}+

^{16}⁄

_{7³}+ ……… —- (i)

Dividing both the sides by 7

^{x}⁄_{7} = ^{1}⁄_{7} + ^{4}⁄_{7²}_{ }+ ^{9}⁄_{7³} +^{ } ………… —- (ii)

Subtracting (ii) from the (i)

^{6x}⁄_{7} = 1 + ^{3}⁄_{7} + ^{5}⁄_{7²} + ^{7}⁄_{7³} + ………… —- (iii)

Obviously the RHS here is an ARITHMETICO-GEOMETRIC series. We, then, divide the equation by 7 again:

^{6x}⁄_{49} = ^{1}⁄_{7} + ^{3}⁄_{7²} + ^{5}⁄_{7³} + ………. —- (iv)

Again subtracting (iv) from (iii)

^{36x}⁄_{49} = 1 + ^{2}⁄_{7} + ^{2}⁄_{7²} + ^{2}⁄_{7³} + ……………

=> ^{36x}⁄_{49} – 1 = ^{2}⁄_{7} + ^{2}⁄_{7²} + ^{2}⁄_{7³} + ……………

Now RHS is a geometric series where, a = ^{2}⁄_{7} and r = ^{2}⁄_{7²} ÷ ^{2}⁄_{7 }= ^{1}⁄_{7}; where a is the first term and r the common ratio.

Formula for sum of infinite terms in a G.P.

S_{∞ }= ^{a}⁄_{1 – r}, if (| r | < 1 )

Therefore ^{36x}⁄_{49} – 1 = ^{2⁄7}⁄_{1-1⁄7}

=> ^{36x}⁄_{49} – 1 =^{2⁄7}⁄_{6⁄7}

=> ^{36x}⁄_{49} – 1 = ^{1}⁄_{3}

=> ^{36x}⁄_{49} = 1/3 + 1

=> ^{36x}⁄_{49 }= 4/3

=> x = 49/27 (option ‘A’)