Given a sequence of 4 numbers, first three of which are in G.P. and the last three are in A.P. with common difference six. If first and last terms of this sequence are equal, then the last term is :

Step-by-step solution

Let a, b, c, d be four numbers of the sequence. Now, according to the question b^2=a c and c-b=6 and a-c=6 Also, given a=d aligned b^2=a c b^2=a [ a+b 2 ] ( 2 c=a+b) aligned a^2-2 b^2+a b=0 Now, c-b=6 and a-c=6 , gives a-b=12 b=a-12 a^2-2 b^2+a b=0 a^2-2(a-12)^2+a(a-12)=0 a^2-2 a^2-288+48 a+a^2-12 a=0 36 a=288 a=8