Stewart's Sequences 4 of 5

$a\times { 2 }^{ n }+b\\ Therefore\quad adding\quad the\quad values\quad of\quad the\quad first\quad two\quad terms:\\ a\times { 2 }^{ 1 }+b=27\quad \\ 2a+b=27\quad ........(1)\\ a\times { 2 }^{ 2 }+b=41\quad \\ 4a+b=41\quad ........(2)\\ Subtracting\quad (1)\quad from\quad (2)\\ 2a=14\\ a=7\\ Putting\quad this\quad back\quad into\quad (1)\\ 2\times 7+b=27\\ b=13\\ Therefore:\\ \frac { \left| 7-13 \right| }{ 7+13 } \\ \frac { 6 }{ 20 } =\boxed { \frac { 3 }{ 10 } }$

You dont even need the values of the next terms!!!

$\quad Let\quad S=27+41+69+125+237+....{ T }_{ n },\quad where\quad { T }_{ n }\quad is\quad the\quad { n }^{ th }\quad term.\\ Also\quad S=\quad \quad \quad 27+41+69+125+......{ T }_{ n-1 }+{ T }_{ n }\\ \\ Subtracting,\quad S-S\quad =\quad 27+\{ (41-27)+(69-41)+....({ T }_{ n }-{ T }_{ n-1 })\} -{ T }_{ n }\\ 0\quad =\quad 27\quad +\quad (14+28+56+....({ T }_{ n }-{ T }_{ n-1 }))\quad -\quad { T }_{ n }\\ \\ Shifting\quad { T }_{ n }\quad to\quad the\quad other\quad side,\\ { T }_{ n }=\quad 27\quad +\quad (14+28+56+....({ T }_{ n }-{ T }_{ n-1 }))\\ \\ Now,\quad the\quad terms\quad in\quad the\quad brackets\quad form\quad a\quad G.P.\quad upto\quad (n-1)\quad terms\\ with\quad first\quad term\quad =\quad 14\quad and\quad common\quad difference\quad =\quad 2.\\ Applying\quad the\quad formula\quad for\quad sum\quad of\quad 'n'\quad terms\quad of\quad a\quad G.P.\\ \\ { T }_{ n }=27\quad +\quad (14\frac { { 2 }^{ n-1 }-1 }{ 2-1 } )\\ { T }_{ n }=27\quad +\quad 14(\frac { { 2 }^{ n } }{ 2 } -1)\quad \quad \quad \quad \quad \quad \quad \quad \{ as\quad { 2 }^{ n-1 }=\frac { { 2 }^{ n } }{ 2 } \} \\ { T }_{ n }=27\quad +\quad 7.{ 2 }^{ n }-14\\ { T }_{ n }=\quad 7.{ 2 }^{ n }+13\\ \\ Comparing,\quad a=7,\quad b=13\\ So,\quad \frac { \left| a-b \right| }{ a+b } =\frac { \left| 7-13 \right| }{ 7+13 } =\frac { 6 }{ 20 } =\frac { 3 }{ 10 }$

CHEERS!!:):)

I just used the fact that we are given the first term and the second term, so we can set up a system of two equations with two unknowns. Solve for a and b, then substitute those values into the given expression.