Question

find the middle term of the sequence formed by all 3 digit number which leave a remainder 3 when divided by 4 also find the sum of all numbers on both sides of the middle temrs separately​

Answer 1

Hey Dear,

◆ Answer -

Middle term = 551

Sum before middle term = 36400

Sum after middle term = 87024

● Explanation -

For the sequence by given conditions.

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

a = 103

l = 999

d = 4

N = 225

Middle term of this sequence will be -

t113 = (a + l) / 2

t113 = (103 + 999) / 2

t113 = 551

Sum of all terms before 551 will be -

S1 = n/2 (a + l)

S1 = 112/2 (103 + 547)

S1 = 36400

Sum of all terms after 551 will be -

S2 = n/2 (a + l)

S2 = 112/2 (555 + 999)

S2 = 87024

Thanks for asking...