Find the middle term of the …

The sequence formed by the given numbers is 103,107,111,115, ...,999.
This is an AP in which a = 103 and d = (107 - 103) = 4.
Let the total number of these terms be n. Then,
T_n = 999 {tex}\Rightarrow{/tex} a + (n-1)d = 999
{tex}\Rightarrow{/tex} 103 + (n -1) {tex}\times{/tex} 4 = 999
{tex}\Rightarrow{/tex} (n-1){tex}\times{/tex}4 = 896 {tex}\Rightarrow{/tex} (n-1) = 224 {tex}\Rightarrow{/tex} n = 225.
{tex}\therefore{/tex} middle term = ({tex}\frac{{n + 1}}{2}{/tex})th term = ({tex}\frac{{225 + 1}}{2}{/tex})th term = 113th term.
T₁₁₃ = (a + 112d) = (103 +112 {tex}\times{/tex} 4) = 551.
{tex}\therefore{/tex} T₁₁₂ = (551 - 4) = 547.
So, we have to find S₁₁₂ and (S₂₂₅ - S₁₁₃).
Using the formula S_m = {tex}\frac{m}{2}{/tex}(a + l) for each sum, we get
s₁₁₂ = {tex}\frac{{112}}{2}{/tex}(103+ 547) = (112{tex}\times{/tex}325) = 36400
(S₂₂₅ - S₁₁₃) = {tex}\frac{{225}}{2}{/tex}(103 + 999) - {tex}\frac{{113}}{2}{/tex}(103 + 551)
= (225 {tex}\times{/tex} 551)-(113 {tex}\times{/tex} 327)
= 123975 - 36951 =87024.
Sum of all numbers on LHS of the middle term is 36400.
Sum of all numbers on RHS of the middle term is 87024.