Question

Find the sum of all natural numbers lying between 100 and 1000, which are multiples of 5.​

Answer 1

Answer:

a¹ (First term) = 100 + 5

                       = 105

aⁿ ( last term ) = 1000 - 5

                       = 995 

d ( Common

      difference) = 5

n ( N0: of terms ) = [(aⁿ - a¹) / d] + 1

                            = (995 -105) /5 ] + 1

                            = (890/5) + 1

                            = 178 + 1

                            = 179//

               ∴  NO: OF TERMS (n) = 179

Xⁿ -------> Sum of all terms 

 

Xⁿ =X / 2 [ a¹ + aⁿ ]

     = 179/2 [ 105 + 995 ]

     = 179/2 ( 1100) 

     = 179 × 550

     = 98450 //

 

         ∴ SUM OF ALL NATURAL NUMBERS ( Xⁿ ) =  98450  

Answer 2

Solution :

we need to find the sum of all natural numbers lying between 100 and 1000 which are multiples of 5.

natural numbers lying between 100 and 1000 which are multiples of 5 are :

105 , 110 , 115 ...... 995

so, here we get a sequence,

110 - 105 = 5

115 - 110 = 5

So, Common difference is same which is 5 so we can say that the given sequence is an AP.

AP : 105 ,110 , 115, 120 .........995

• First term (a) = 105
• common difference (d) = 5
• last term = 995
• number of terms = ??
• sum of all natural numbers = ??

we know that,

>> an = a + (n - 1)d

>> 995 = 105 + (n - 1)5

>> 995 - 105 = (n - 1)5

>> 890/5 = n - 1

>> 178 + 1 = n

>> n = 179

• Number of terms in AP = 179

Sum of all natural numbers which lies b/w 100 - 1000 which are multiples of 5 are :

sn = n/2[a + l]

sn = 179/2[105 + 995]

sn = 179/2 × 1100

sn = 179 × 550

sn = 98450

Hence,

• sum of all natural numbers lying between 100 and 1000, which are multiples of 5 is 98450

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