Math, asked by shajisaifina2632, 9 months ago

Find the sum of all natural numbers between 250 and 1000 which are exacty divisible by 7

Answers

Answered by shashvatmalik3907
1

Answer:

The numbers between 250 and 1000 which are divisible by 3 are 252, 255, 258, ......,999

This is an A.P. whose first term a = 252, Common difference, d = 3 and Last term l=999

We Know that

l=a+(n-1)d

999=252+(n-1)3

999 - 252 = 3(n - 1)

n=250

We know that

S=n/2(a+l)

=250/2(252+999)

=125*1251

=156375

Hence, the required sum is 156375.(Ans)

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Step-by-step explanation:

Answered by snehamittal121
0

step by step explanation:-

we have numbers 250 to 1000

we can do it by arthmetic progressions

check which one no. is divisible by 7 ahead 250

i.e. 252

And now, do it again from the last one which no. divisible by before 1000 i.e. 994

now we can apply formula i.e. an=a+(n-1) ×d

where an= last term of an AP

a= first term of an AP

n= no. of terms in an AP

d= common difference of an AP

an= a+(n-1)×d

994= 252+(n-1)×7

994-252= (n-1)×7

742/7 = n-1

106 =n-1

106+1 = n

so, n = 107

now we can find their sum

by following formula

Sn = n/2{2a+(n-1)×d}

Sn= 107/2 { 2× 252 +( 107-1) ×7}

Sn= 107/2 {504 + (106×7)}

Sn= 107/2 (504+742)

Sn=107/2(1246)

Sn= 107/2 × 1246

Sn= 107× 623

Sn= 66661

so, the sum of all natural numbers between 250 to 1000 which are exactly divisible by 7 is 66661

hey mate here's ur answer with explanation hope it helps

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