### Solution:

Given pair of linear equations $\n \n 2x+y\u22125=0\n $ and $\n \n (a+b)x+(5a\u22127b)y\u221220=0\n $ has infinite number of solutions. We need to find the value of $\n \n a&b\n $ .A system of linear equations $\n \n \n a\n 1\n \n x+\n b\n 1\n \n y+\n c\n 1\n \n =0\n $ and $\n \n \n a\n 2\n \n x+\n b\n 2\n \n y+\n c\n 2\n \n =0\n $ will have infinite solutions when $\n \n \n \n \n a\n 1\n \n \n \n \n a\n 2\n \n \n \n =\n \n \n b\n 1\n \n \n \n \n b\n 2\n \n \n \n =\n \n \n c\n 1\n \n \n \n \n c\n 2\n \n \n \n \n $ .

Now from $\n \n 2x+y\u22125=0\n $ and $\n \n (a+b)x+(5a\u22127b)y\u221220=0\n $ , we have,

$\n \n \n a\n 1\n \n =2,\n b\n 1\n \n =1,\n c\n 1\n \n =\u22125\n $ and $\n \n \n a\n 2\n \n =a+b,\n b\n 2\n \n =5a\u22127b,\n c\n 2\n \n =\u221220\n $

Now, substituting the values in the condition we have,

$\n \n \n 2\n \n a+b\n \n =\n 1\n \n 5a\u22127b\n \n =\n \n \u22125\n \n \u221220\n \n \n $

On comparing, we have,

$\n \n \n \n \n 2\n \n a+b\n \n =\n 5\n \n 20\n \n \n \n \n \n \n \u21d22(20)=5(a+b)\n \n \n \n \n \u21d240=5(a+b)\n \n \n \n \n \u21d2a+b=\n \n 40\n 5\n \n \n \n \n \n \n \u21d2a+b=8\u2009\u2009\u2009...(i)\n \n \n \n \n $

And,

$\n \n \n 1\n \n 5a\u22127b\n \n =\n \n \u22125\n \n \u221220\n \n \n $

$\n \n \n \n \u21d220=5(5a\u22127b)\n \n \n \n \n \u21d25a\u22127b=\n \n 20\n 5\n \n \n \n \n \n \n $

$\n \n \u21d25a\u22127b=4\u2009\u2009\u2009...(ii)\n $

Multiply equation (i), by 7, we have,

$\n \n 7a+7b=56\n $

Now adding the above equation to equation (ii),

$\n \n \n \n 7a+7b+5a\u22127b=56+4\n \n \n \n \n \u21d212a+0=60\n \n \n \n \n \u21d212a=60\n \n \n \n \n \u21d2a=\n \n 60\n \n 12\n \n \n \n \n \n \n \u21d2a=5\n \n \n \n \n $

Now, substituting the value of $\n a\n $ in equation (i), we have,

$\n \n \n \n 5+b=8\n \n \n \n \n \u21d2b=8\u22125\n \n \n \n \n \u21d2b=3\n \n \n \n \n $

Thus, the value of $\n a\n $ is 5 and the value of $\n b\n $ is 3.

Therefore, option 2 is correct.