If the first and the (2n+1)th terms of an A.P., G.P. and H.P. are equal and their nth terms are respectively a, b, c then always

Question

If the first and the (2n+1)th terms of an A.P., G.P. and H.P. are equal and their nth terms are respectively a, b, c then always (A) a = b = c (B) a ge b ge c (C) a + c = b (D) ac - b2 = 0

Tardigrade Admin · Accepted Answer

There seems to be a printing mistake here If there are (2n-1) terms instead of (2n + 1) terms then nth terms of the A.P., G.P. and H.P. are the A.M., G.M. H.M of the first and the last terms. So, a ge b ge c ac-b2(B, D)
otherwise if there are (2n + 1) terms then the nth terms should be in decreasing order of
A.P., G.P. H.P. i. e.a ge b ge c (B)

Q. If the first and the $(2 n + 1)^{t h}$ terms of an $A . P ., G . P .$ and $H . P .$ are equal and their $n^{t h}$ terms are respectively $a, b, c$ then always

$a = b = c$

$a \geq b \geq c$

$a + c = b$

$a c - b^{2} = 0$

Q. If the first and the (2n+1)th terms of an A.P.,G.P. and H.P. are equal and their nth terms are respectively a,b,c then always

a=b=c

a≥b≥c

a+c=b

ac−b2=0

Q. If the first and the $(2 n + 1)^{t h}$ terms of an $A . P ., G . P .$ and $H . P .$ are equal and their $n^{t h}$ terms are respectively $a, b, c$ then always

$a = b = c$

$a \geq b \geq c$

$a + c = b$

$a c - b^{2} = 0$