# find the middle term of the sequence formed by all three digits numbers which leave a reminder 3, when divided by 4. Also find the sum of all numbers on both sides of the middle term separately.

103 , 107 , 111 , 115 , .... 999

The above list is in AP with first term, a = 103 and common difference, d = 4

Let n be the number of terms in the AP.

Now, a

_{n }= 999

103 + ( n - 1 ) 4 = 999

103 + 4

*n*- 4 = 999

4

*n*+ 99 = 999

4

*n*= 900

*n*= 225

Since, the number of terms is odd, so there will be only one middle term.

$\mathrm{middle}\mathrm{term}={\left(\frac{\mathrm{n}+1}{2}\right)}^{\mathrm{th}}\mathrm{term}={113}^{\mathrm{th}}\mathrm{term}=\mathrm{a}+112\mathrm{d}=103+112\times 4=551\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{We}\mathrm{know}\mathrm{that},\mathrm{sum}\mathrm{of}\mathrm{first}\mathrm{n}\mathrm{terms}\mathrm{of}\mathrm{an}\mathrm{AP}\mathrm{is},\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{n}}{2}\left[2\mathrm{a}+\left(\mathrm{n}-1\right)\mathrm{d}\right]\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Now},{\mathrm{S}}_{112}=\frac{112}{2}\left[2\times 103+111\times 4\right]=36400\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Sum}\mathrm{of}\mathrm{all}\mathrm{terms}\mathrm{before}\mathrm{middle}\mathrm{term}={\mathrm{S}}_{112}=36400\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{\mathrm{S}}_{225}=\frac{225}{2}\left[2\times 103+224\times 4\right]=123975\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Now},\mathrm{sum}\mathrm{of}\mathrm{terms}\mathrm{after}\mathrm{middle}\mathrm{term}={\mathrm{S}}_{225}-\left({\mathrm{S}}_{112}+551\right)=123975-\left(36400+551\right)=87024\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

**
**