A elastic spring has length l1 when tension in it is4N . Its length l2 when tension in it is5N .Its length l3 when tension in it is 9N. Find the sum of original length  and l3?

Let ‘l ’ be the natural length(l1−l)∝4----(1)(l2−l)∝5-----(2)(l3−l)∝9----(3)dividing (1) by (2) we get (l1−l)(l2−l)=45⇒5l1-4l2=l---(4)dividing (1) by (3) we get (l1−l)(l3−l)=49---(5)from (4) and (5) we get ⇒  (l1− 5l1-4l2)(l3− 5l1-4l2)=49on solving we get l3=5l2-4l1---(6)l+l3=5l1-4l2+5l2-4l1=l1+l2