Correct option is (B) 128
The first number which is divisible by 7, between 100 & 1000 is 105 and last number which is divisible by 7 between 100 & 1000 is 994.
\(\therefore a_1=105,d=7\;\&\;a_n=994\)
\(\Rightarrow a+(n-1)d=994\) \((\because a_n=a+(n-1)d)\)
\(\Rightarrow105+(n-1)7=994\)
\(\Rightarrow7(n-1)=994-105=889\)
\(\Rightarrow n-1=\frac{889}7=127\)
\(\Rightarrow n=1+127=128\)
Hence, total 128 numbers between 100 & 1000 which are divisible by 7.
