find the middle term of the …

The three digit numbers which leaves 3 as remainder when divided by 4 are : 103,107, 111,..... 999
Now, the first number{tex} a = 103{/tex}, last number l = 999
and common difference d = 4
Let the number of terms in this sequence be {tex}n{/tex}.
{tex}l = a + (n - 1)d{/tex}
or, {tex}103 + ( n - 1)4 = 999{/tex}
(n -1)4 = 999 - 103
{tex}(n -1)4 = 896{/tex}
{tex}(n - 1) {/tex}= {tex}\frac{896}{4}{/tex}= 224
{tex}n = 224 + 1 = 225{/tex}
Middle term = {tex}\frac { 225 + 1 } { 2 }{/tex}
= 113^th term
a₁₁₃ = 103 + 112 {tex}\times{/tex} 4 = 551
and a₁₁₂ = 551 - 4 = 547
Sum of first 112 terms
{tex}= \frac { 112 } { 2 } ( 103 \times 2 + 111 \times 4 ){/tex}
{tex}= \frac { 112 } { 2 } ( 206 + 444 ){/tex}
{tex}= 56 \times 650{/tex}
= {tex}36400{/tex}
and a₁₀₄ ={tex} 551 + 4 = 555{/tex}
The Sum of last 112 terms
={tex}\frac { 112 } { 2 } ( 2 \times 555 + 111 \times 4 ){/tex}
{tex}= 56 ( 1110 \times 444 ){/tex}
{tex}= 56 \times 1554{/tex}
= 87024