Question

Question is find the middle term of the sequence formed by all the all the digit number which leaves remainder 3 when divided by 4 also find the sum of all number on but both side of the middle term sequence

Answer 1

Answer:

36400

87024

Step-by-step explanation:

first term 103 = 25*4 + 3

last term 999 = 249*4 + 3

d = 4

a= 103

nth term = a + (n-1)d

999 = 103+ (n-1)4

n-1 = 896/4

n-1 = 224

n = 225

total 225 terms

middle term = (225+1)/2 = 113th Term

Total sum = (225/2)(103+999) = 123975

112th Term = 103 + (112-1)4 = 547

Sum of first 112 terms = (112/2)(103 + 547)

= 36400

Last 112 terms

114th term  to 225 terms

114 term = 103 + (114-1)4 = 555

114th term now become 1st term

225th term become 112th term and last term = 999

Sum of last 112 terms

=(112/2)(555+999)

= 87024