Question

A sequence a,ax,ax^2,______ax^n, has odd no. of terms.Then median is..

1. ax^n
2.ax^n/2
3.ax^n/2-1
4.ax^n/2+1

Answer 1

a sequence : a, ax , ax² , ax³ , ..... axⁿ has odd number of terms.

we can see that a, ax , ax² , ax³ ..... axⁿ are in geometric progression. because ratio of two consecutive terms is always same.
e.g., ax/a = ax²/ax = ax³/ax² = ...... axⁿ/axⁿ-¹ = x

again come to the question,
question want to find the median,
if number of terms is an odd number.
Then, median =$\left(\frac{N+1}{2}\right)th$ term in ascending order of sequence.
here number of terms in sequence is (n + 1).
so, N = (n + 1)

then, we have to find

so, find we have to find $\left(\frac{n+2}{2}\right)th$ term of sequence .

here, first term,$a_1$ = a
common ratio, r = x
we know nth term of GP is given by
$T_n=a_1r^{n-1}$

so, $\left(\frac{n+2}{2}\right)th$ of sequence is
$T_{\left(\frac{n+2}{2}\right)}=ax^{\left(\frac{n+2}{2}\right)-1}$

=$ax^{n/2}$

Answer 2

For the sequence a, ax, ax^2, ….., ax^n which have the odd number of terms in total. Then, the median in such situation will be ax^n/2.

Since the total terms are odd n+1 in number, the median will come out to be even that is n will be even.

For example, if there are a total of 5 terms then the third term will be the median.