JEE Advanced 2015 Problem

$We\quad know,\quad \frac { { S }_{ 7 } }{ { S }_{ 11 } } \quad =\quad \frac { 6 }{ 11 } \\ =>\quad \frac { \frac { 7 }{ 2 } [2a\quad +\quad 6d]\quad }{ \frac { 11 }{ 2 } [2a\quad +\quad 10d] } \quad =\quad \frac { 6 }{ 11 } \\ =>\quad 77[2a\quad +\quad 6d]\quad =\quad 66[2a\quad +\quad 10d]\\ =>\quad \quad 7[2a\quad +\quad 6d]\quad =\quad 6[2a\quad +\quad 10d]\\ =>\quad 14a\quad +\quad 42d\quad =\quad 12a\quad +\quad 60d\\ =>\quad a\quad =\quad 9d\\ \\ \quad \therefore \quad { 7 }^{ th }\quad term\quad =\quad { a }_{ 7 }\quad =\quad a\quad +\quad 6d\quad =15d\\ \\ \because \quad 130\quad <{ \quad 7 }^{ th }\quad term\quad <\quad 140\\ =>130\quad <\quad 15d\quad <\quad 140\\ \\ \therefore \quad 15d\quad =\quad 135\\ =>\quad d\quad =\quad 9$

What manish did I did the same

https://brilliant.org/problems/can-you-figure-it-out/?group=L9XW7iIU36z4&ref_id=802119 Same problem

Same method as Manish Dash one's.Was a easy one but really happy to solve it.After all a JEE ADVANCE one.