If the (a+1)th ,7th ,and (b+1)th terms of an A.P, are in G.P with a,6,b being in H.P, then the 4th term of this A.P is ? (A) -7/2 (B) 7/2 (C) 0 (D) 3

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Answer 1 · 2018-04-29T07:20:42

(C) 0.

For the AP in Question, suppose the #n^(th)# term and

the common difference are , resp., #t_n and d!=0#.

If #t_1=x," then, "t_n=x+(n-1)d, n in NN#.

Given that, #t_(a+1), t_7, t_(b+1)# are in GP.

#:. (x+ad), (x+6d), (x+bd)# are in GP.

#:. (x+6d)^2=(x+ad)(x+bd)#.

#:. cancel(x^2)+12xd+36d^2=cancel(x^2)+(a+b)xd+abd^2#.

Dividing by #d!=0, 12x+36d=(a+b)x+abd......(star)#.

Here, #a,6,b# are in HP, or, #1/a, 1/6, 1/b# in B

#:. 2*(1/6)=1/a+1/b, i.e., 1/3=(a+b)/(ab)#.

# :. a+b=1/3ab#.

Sub.ing in #(star), 12x+36d=1/3abx+abd=1/3ab(x+3d)#

#:. 12(x+3d)-1/3ab(x+3d)=0#

#:. (x+3d)(12-1/3ab)=0#.

#:. (x+3d)=0, or, ab=36#.

In case, #ab=36," then "a+b=1/3ab=12#.

Solving these for #a and b#, we get, #a=b=6#.

But, in that event, #t_(a+1)=t_(b+1)=t_7# will make #d=0#.

So, let us discard that possibility.

We are now left with #x+3d=0#, which is, in fact, #t_4#.

Hence, the Right Option is (C) 0.

Enjoy Maths!