Question

write the sequence of all three digit numbers which leaves remainder 3 on division by 7 which is last term of this sequence​

Answer 1

Step-by-step explanation:

• first three digit number is 100
• if 100 divides by 7 then remainder is 2
• so 101 is the first number who when divides by 7 remaining 3
• next number must be greater by 7

• so it is 108

• so series is

101 , 108 , 115 ,.........

• last three digit number is 999 when divides by 7 remaining 5

• so 997 is the number when divides by 7 remaining 3

• so series is

101 , 108 , 115 ,......... ,997

here a = 101

d = 7

tn = 997

n = ?

tn = a + (n-1) d

997 = 101 + ( n - 1 ) × 7

997 - 101 = ( n - 1 ) × 7

( n - 1 ) × 7 = 896

n - 1 = 896 / 7

n - 1 = 128

n = 128 + 1

n = 129

so 129 numbers

• If 100 divides by 7 then remainder is 2

• so 101 is the first number who when divides by 7 remaining 3

• next number must be greater by 7

• so it is 108

• so series is

101 , 108 , 115 ,.........

• last three digit number is 999 when divides by 7 remaining 5

• so 997 is the number when divides by 7 remaining 3

• so series is

101 , 108 , 115 ,......... ,997

here a = 101

d = 7

tn = 997

n = ?

tn = a + (n-1) d

997 = 101 + ( n - 1 ) × 7

997 - 101 = ( n - 1 ) × 7

( n - 1 ) × 7 = 896

n - 1 = 896 / 7

n - 1 = 128

n = 128 + 1

n = 129

answer:- 129 numbers