Question

Find the sum of a) all three digit natural numbers, which are divisible by 7. b) all odd integers between 2 and 100 divisible by 3. c) the first 40 positive integers divisible by 6 d) first n odd natural numbers. e) all integers between 84 and 719, which are multiples of 5 [this is a question from arithmetic progressions. please answer in detail} VERY URGENT KINDLY HELP

Answer 1

Answer:

Step-by-step explanation:

→All three-digit natural numbers which are divisible by 7 are

105,112,119,126,...,994 which are in AP with a=105,d=7,l=994

Let the number of these terms be n. Then,

$a_{n}$=994

a+(n−1)d=994

105+(n−1)×7=994

n=128

Now, $S_{n}$=$\frac{n}{2}$[a+l]

​        =$\frac{128}{2}$ (105+994)=70336.

→The odd integers between 2 and 100 which are divisible by 3 are 3, 9, 15, 21, ..., 99.

Here, a=3 and d=6.

$a_{n}$=99

⇒a+(n−1)d=99

⇒3+(n−1)×6=99

⇒6(n−1)=96

⇒n−1=16

⇒n=17

Therefore, Required sum = $S_{n}$=$\frac{n}{2}$[a+l]

                                        =$\frac{17}{3}$[3+99]  

                                           =(3+99)=867

→the first 40 positive integers divisible by 6

We know that the first 40 positive integers divisible by 6 are 6,12,18,....

This is an AP with a = 6 and d = 6

                   $S_{n} =\frac{n}{2} [2a+(n-1)d]$

$S_{40}$= 20[2(6) +(40-1)6] =20[12+234] =4920

→first n odd natural numbers.

a=1 , d = 2

Let, last term, l = (2n-1)

Sn = ( n/2) × (a+l )

Sn = (n/2) × (1 + 2n – 1)

Sn = (n/2) × (2n)

       = $n^{2}$

→All integers between 84 and 719, which are multiple of 5 are

85,90,95,......715

which forms an A.P

first term of this A.P is $a_{1}$ =85

second term of this A.P is $a_{2}$=90

last term of this A.P is $a_{n}$=715

common difference  

d=$a_{2}$-  $a_{1}$

d=90−85=5 .

nth term of this A.P is given by

$a_{n}$=$a_{1}$+(n−1)d

$a_{n}$=715;$a_{1}$=85 and d=5 in above equation we get,

⟹715=85+(n−1)5

⟹5n−5+85=715

⟹5n=715−80=635

n=$\frac{635}{5}$  =127 number of terms in this A.P

now, sum of these n=127 terms is given by

$S_{n}$=$\frac{n}{2}$[$a_{1}$+$a_{n}$]

put values

$S_{127} =\frac{127}{2} X800$

$S_{127} =127X400$

$S_{127} =50800$

hence the sum of all integers between 84 and  719 which are multiple of 5 , is $S_{127} =50800$

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