Question -
Answer -
┬аFirst three-digit number that is divisible by7 are;
First number = 105
Second number = 105+7 = 112
Third number = 112+7 =119
Therefore, 105, 112, 119, тАж
All are three digit numbers are divisible by 7and thus, all these are terms of an A.P. having first term as 105 and commondifference as 7.
As we know, the largest possible three-digitnumber is 999.
When we divide 999 by 7, the remainder will be5.
Therefore, 999-5 = 994 is the maximum possiblethree-digit number that is divisible by 7.
Now the series is as follows.
105, 112, 119, тАж, 994
Let 994 be the┬аnth term of this A.P.
first term, a┬а= 105
common difference, d┬а= 7
an┬а= 994
n┬а= ?
As we know,
an┬а=┬аa+(nтИТ1)d
994 = 105+(nтИТ1)7
889 = (nтИТ1)7
(nтИТ1) = 127
n┬а= 128
Therefore, 128 three-digitnumbers are divisible by 7.