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Question

Mathematics Question on Sequence and series

If the first and the (2n1)th (2n - 1)th term of an AP,GPAP, GP and HPHP are equal and their nth terms are a,ba, b and cc respectively, then

A

a=b=ca = b = c

B

abca \ge b \ge c

C

a+c=ba + c = b

D

acb2=0ac - b^2 =0

Answer

acb2=0ac - b^2 =0

Explanation

Solution

Since, first and (2n - 1)th terms are equal.
Let first term be x and (2n - 1)th term be y,
whose middle term is tn.t_n.
Thus, in arithmetic progression, tn=x+y2=at_n =\frac{x+y}{2} = a
In geometric progression, tn=xy=bt_n =\sqrt {xy} = b
In harmonic progression, tn=2xyx+y=ct_n =\frac{2xy}{x+y} =c
b2=ac\Rightarrow \, \, \, b^2 = ac and a>b>c[usingAM>GM>HM]a > b >c \, \, \, \, \, \, [using\, AM > GM > HM]
Here, equality holds (i.e. a = b =c) only if all terms are
same. Hence, options (a), (b) and (d) are correct