Question

Find the sum of all 3 digit number which are multiple of 7.

Answer 1

Step-by-step explanation:

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Answer 2

Solution:-

Let •a=first term of Arithematic progression

•d=common difference of Arithematic progression

•Tn=nth term of Arithematic progression

•l=last term of Arithematic progression

•Sn=sum of nth term of Arithematic progression

Formula used:-

•Tn=a+(n-1)d

•Sn=n/2(a+l)

Calculation:-

All 3 digit number which are multiple of 7 are

105,112,119,126,.......994

This forms an A•P ,where a=105,d=(112-105)=7 and l=994

Let the given AP contain n terms ,Then ,

Tn=994 =>a+(n-1)d=994

=>105+(n-1)×7=994

=>(n-1)×7=889

=>(n-1)=127

=>n=128

•Required sum =n/2(a+l)

=128/2×(105+994)=(64×1099)

=64×(1100-1)

=(64×1100)-(64×1)

=(70400-64)=70336

•The sum of all three digit numbers which are multiple of 7 is 70336.