∵a,b,c are in A.P. then a+c=2b also it is given that, a+b+c=43⇒2b+b=43⇒b=41 Again it is given that, a2,b2,c2 are in G.P. then (b2)2=a2c2⇒ac=±161 From (1), (2) and (3), we get; a±16a1=21⇒16a2−8a±1=0 Case I: 16a2−8a+1=0 ⇒a=41 (not possible as a<b) Case II: 16a2−8a−1=0⇒a=328±128 ⇒a=41±221∴a=41−221(∵a<b)