Soppose that all the terms of an arithmetic progression (AP) are natural numbers. If the ratio of the sum of the first seven terms to the sum of the first eleven terms is `6:11` and the seventh term lies between 130 and 140, then the common difference of this AP is

Explanation

Let first term = a and common difference = d
`because" "("sum of seven terms")/("sum of eleven terms")=6/11`
` rArr" "(7/2 (a_1+a_7))/(11/2(a_1+a_11))=6/11rArr(7/2(2a+6d))/(11/2(2a+10d))=6/11`
or `a=9d` and `130 lt a_7 lt 140`
`rArr" "130 lt a_1 +6d lt 140 rArr 130 lt 15d lt 140`
`:." "8""2/3lt d lt 9""1/3" "rArr" "d=9" "(because a, d in N)`