Three-digit numbers divisible by 7 form an A.P.
First three-digit multiple of 7: ( 105 ) (( 7 \times 15 ))
Last three-digit multiple of 7: ( 994 ) (( 7 \times 142 ))
[ a = 105,\ d = 7,\ a_n = 994 ]
[ 105 + (n-1)7 = 994 ] [ (n-1)7 = 889 ] [ n-1 = 127 ] [ n = 128 ]
Answer: 128