Question

Find the middle term of the sequence formed by all three digit number which leaves a remainder 3 when divided by 4 . Also fimd the sum of all numbers on both side of the middle term separately

Answer 1

$\bf{\huge{\underline{\underline {\mathfrak{\tt{\blue{Answer}}}}}}}$

The sequence of the three digit numbers which leaves a remainder 3 when divided by 4 are :

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

(Note : Be in your conscious while reading the question. Don't move to 4 digit number)

First term = a = $\bf {t_1}$ = 103

$\bf{t_2}$ = 107

$\bf{t_3}$ = 111

$\bf{t_3}$ = 115

$\bf{t_n}$ = 999

Common difference, d = $\bf{t_2}$ - $\bf{t_1}$

= 107 -103

= 4

Common difference, d = $\bf{t_3}$ - $\bf{t_2}$

= 107 - 111

= 4

Common difference d is constant.

•°• the sequence is an A. P

We know the formula to find the $\bf{n_th}$ of an A. P,

$\bf{t_n}$ = a + (n - 1) d

999 = 103 + (n-1) 4

999 = 103 + 4n - 4

999 = 99 + 4n

999 - 99 = 4n

900 = 4n

n = $\bf\frac{900}{4}$

$\bf{\boxed{\red{n = 225}}}$

To find : Middle term of the sequence.

As the middle term is an odd number, 225 there would be only one middle term.

Middle term = $\bf\frac{a +1}{2}$

Middle term = $\bf\frac{225+1}{2}$

Middle term = $\bf\frac{226}{2}$

Middle term = 113th.

$\bf{t_1_1_3}$ = a + (n - 1) d

$\bf{t_1_1_3}$ = 103 + (113 - 1) 4

$\bf{t_1_1_3}$ = 103 + 112 × 4

$\bf{t_1_1_3}$ = 103 + 448

$\bf{t_1_1_3}$ = 551

•°•Middle term = 551

To find : The sum of all numbers on both sides of the middle terms separately , i. e the sum of numbers before and after $\bf{t_1_1_2}$

Case 1: $\bf{t_1}$ to $\bf{t_1_1_2}$

$\bf{t_1}$ = 103

Difference, d = 4

$\bf{t_1_1_2}$ = 551 - 4 = 547

$\bf{S_n}$ = $\bf\frac{n}{2}$($\bf{t_1}$ + $\bf{t_1_1_2}$ )

$\bf{S_1_1_2}$ = $\bf\frac{112}{2} [tex]\bf{103}$ + $\bf{547}$

$\bf{S_1_1_2}$ = 56 × 650

$\bf{S_1_1_2}$ = 36400

Sum of all numbers after $\bf{t_1_1_2}$ i.e from $\bf{t_1_1_4}$ to $\bf{t_2_2_5}$

$\bf{t_114}$ = 551 + 4 = 555

d = 4

$\bf{t_2_2_5}$ = 999

$\bf{S_2_2_5- 1_1_3}$ = $\bf\frac{n}{2}$ ($\bf{t_114}$ + $\bf{t_225}$ )

$\bf{S_2_2_5-1_1_3}$ = $\bf\frac{225-113}{2}$ ($\bf{555}$ + $\bf{999}$ )

$\bf{S_2_2_5-_1_1_3}$ = $\bf\frac{112}{2} × [tex]\bf{555}$ + $\bf{999}$

$\bf{S_2_2_5-_1_1_3}$ = 56 × 1554

$\bf{S_2_2_5-_1_1_3}$=

87024