Question
Find the middle term of sequence formed by all three digit numbers which leave a remainder $3$ when divided by $4.$ Also find sum of all numbers on both sides of the middle term.
✓
Answer
The sequence formed by the given numbers is $103, 107, 111, 115, .... , 999.$
This is an $A.P.$ with $a=103, d=107-103=4$ and
$t_{ n }=999$
$t_n=a+(n-1) d$
$\therefore 999=103+(n-1) 4$
$\therefore 896=4 n-4$
$\therefore 4 n=900$
$\therefore n=\frac{900}{4}$
$\therefore n=225$
$ \therefore \text { Middle term } =\left(\frac{n+1}{2}\right)^{ th } \text { term }$
$ =\left(\frac{225+1}{2}\right)^{ th } \text { term }$
$ =\frac{226}{2}$
$ =113^{ th } \text { term. }$
$t_n=a+(n-1) d$
$\therefore t_{113}=103+(113-1) 4$
$\therefore \quad=103+112 \times 4$
$=103+448$
$t_{113}=551$
Now, $t_{112}=551-4=547$
So, we have to find $S_{112}$ and $\left(S_{225}-S_{113}\right)$
$S _n=\frac{n}{2}\left[t_1+t_{ n }\right]$
$\therefore S _{112}=\frac{112}{2}[103+547]$
$\therefore \quad=\frac{112}{2} \times 650$
$\therefore S _{112}=36400$
Similarly,
$ S _{225}- S _{113} =\frac{225}{2}(103+999)-\frac{113}{2}(103+551)$
$ =(225 \times 551)-(113 \times 327)$
$ =123975-36951$
$ =87024$
$\therefore$ Sum of all members on $\text{LHS}$ of middle term is $36400$ and sum of all numbers on $\text{RHS}$ of the middle term is $87024 .$
This is an $A.P.$ with $a=103, d=107-103=4$ and
$t_{ n }=999$
$t_n=a+(n-1) d$
$\therefore 999=103+(n-1) 4$
$\therefore 896=4 n-4$
$\therefore 4 n=900$
$\therefore n=\frac{900}{4}$
$\therefore n=225$
$ \therefore \text { Middle term } =\left(\frac{n+1}{2}\right)^{ th } \text { term }$
$ =\left(\frac{225+1}{2}\right)^{ th } \text { term }$
$ =\frac{226}{2}$
$ =113^{ th } \text { term. }$
$t_n=a+(n-1) d$
$\therefore t_{113}=103+(113-1) 4$
$\therefore \quad=103+112 \times 4$
$=103+448$
$t_{113}=551$
Now, $t_{112}=551-4=547$
So, we have to find $S_{112}$ and $\left(S_{225}-S_{113}\right)$
$S _n=\frac{n}{2}\left[t_1+t_{ n }\right]$
$\therefore S _{112}=\frac{112}{2}[103+547]$
$\therefore \quad=\frac{112}{2} \times 650$
$\therefore S _{112}=36400$
Similarly,
$ S _{225}- S _{113} =\frac{225}{2}(103+999)-\frac{113}{2}(103+551)$
$ =(225 \times 551)-(113 \times 327)$
$ =123975-36951$
$ =87024$
$\therefore$ Sum of all members on $\text{LHS}$ of middle term is $36400$ and sum of all numbers on $\text{RHS}$ of the middle term is $87024 .$
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