e lirst ree lens.How many 3-digit numbers are divisible by 7 ?

1.	e lirst ree lens.How many 3-digit numbers are divisible by 7 ?
Answer» first = 105last = 994so , Tn = a+(n-1)dHere , 994 =105+(n-1)7or , 7(n-1)=994-105or , 7(n-1)=889or , n-1=127or , n=128 . 3 digit no are 100 to 999a= 104d=7ln=994let n be the term a+(n-1)d=on104+(n-1)×7=9947n-7=994-104=8897n= 889+7=896n= 896/7=128total no of term = 128 Numbers begin from 105 ....... 994 here a= 105l= 994n=?d= 7so994 = 105+(n-1) 7= 105+7n-7994= 98+7n994-98 = 7n896/7 = nn= 128 So , 128 three digit numbers are divisible by 7 Least 3 digit number divisible by 7 is 105The biggest 3 digit number divisible by 7 is 994 Sn= a+(n-1)d=>994= 105+(n-1)7=>7(n-1)= 889 {994-105=889}=>n-1= 127=> n=128... The first number =105,last number=994,last term= a+(n-1)d,994=105+(n-1)7,889=(n-1)7,127=n-1,n=126

Answer»

first = 105last = 994so , Tn = a+(n-1)dHere , 994 =105+(n-1)7or , 7(n-1)=994-105or , 7(n-1)=889or , n-1=127or , n=128 .

3 digit no are 100 to 999a= 104d=7ln=994let n be the term a+(n-1)d=on104+(n-1)×7=9947n-7=994-104=8897n= 889+7=896n= 896/7=128total no of term = 128

Numbers begin from 105 ....... 994

here a= 105l= 994n=?d= 7so994 = 105+(n-1) 7= 105+7n-7994= 98+7n994-98 = 7n896/7 = nn= 128

So , 128 three digit numbers are divisible by 7

Least 3 digit number divisible by 7 is 105The biggest 3 digit number divisible by 7 is 994

Sn= a+(n-1)d=>994= 105+(n-1)7=>7(n-1)= 889 {994-105=889}=>n-1= 127=> n=128...

The first number =105,last number=994,last term= a+(n-1)d,994=105+(n-1)7,889=(n-1)7,127=n-1,n=126

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