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