Question

Find the middle term of the sequence formed by all the numbers between $9$ and $95$ which leave a remainder $1$ when divided by $3$. Also find the sum of numbers on both the sides of the middle term separately.

Answer 1

Here, the required numbers are,

$\Rightarrow 10,13,16,......,94$

Here,

$a=10$

$d=3$

$l=94$

Then,

$l=a+(n-1)d$

$94=10+(n-1)3$

$3n-3=84$

$n=\frac{87}{3}$

$n=29$

Here, the number of terms in the AP is odd. So, it has only one middle term and it will be ${15}^{th}$ term.

So,

$a_{15}=10+(15-1)3$

$a_{15}=52$

Now, sum of first 14 terms is,

$S_{14}=\frac{14}{2}[2\times 10+(14-1)3]$

$S_{14}=\frac{14}{2}\left[59\right]$

$S_{14}=413$

Again, sum of last 14 terms is,

$\Rightarrow S_{29}-[S_{14}+a_{15}]$

$\Rightarrow \frac{29}{2}[10+84]-(413+52)$

$\Rightarrow 1508-465$

$\Rightarrow 1043$

Hence, the required middle term is $52$, the sum of first half terms is $413$ and the sum of last half terms is $1043$.