Hi,
My two cents for this question.
So given series is arithmetic series where say first term is \(a_1\) and last term is \(a_n\).
We know that AP is given by \(a_n\)= \(a_1\) +(n-1)d----(1) where n is number of terms and d is common difference .
Also if we recall that in AP we can find mean = \(\frac{a_1+a_n}{2}\).
It is always true of evenly spaced sets that the median – the middle term of the set – is equal to the mean, or arithmetic average, of the set.
So in AP mean = median.
Let median M, and Mean =m
M=m= \(\frac{a_1+a_n}{2}\)
so \(a_1\)= 2m-\(a_n\) which is also same as 2M-\(a_n\) -------(2)( Becuase M=m)
substituting 2 in 1 we get
\(a_n\) =2M-\(a_n\) +(n-1)d
Now since the given series is of consecutive of integers we have d =1
so \(a_n\) =2M-\(a_n\) +(n-1)
\(2a_n\)=2M +(n-1)
\(a_n\)=M +\(\frac{n-1}{2}\)
Here we are given that M=120 and n= Q
So we have
\(a_n\)= 120 +\(\frac{Q-1}{2}\)
Probably logic behind this question is above .
Probus
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Probus
~You Just Can't beat the person who never gives up~ Babe Ruth