Suppose that all terms of an arithmetic progression (A.P) are natural numbers.If the reation of the sum of the first sece terms to the sum of the firest eleven terms is 6 : 11 and the seventh term lies in between 130 and 140, then the common difference of this A.P is

Options

A. 5
B. 6
C. 8
D. 9

Answer

C

Explanation

Let the term and common difference of the A.P. be a and respectively. It is given that
`(S_(7))/(S_(11))=(6)/(11)rArr((7)/(2){2a+6d})/((11)/(2){2a+10d})=(6)/(11)rArr(7(a+3d))/((a+5d))=6rArra=9d`
It is also given that
`rArr" "130lta_(7)lt140`
`rArr" "130lt9d+6dlt140rArr130lt15dlt140rArrd=9`.