pls answer this nmtc question with steps 11+192+1993+19994+........+1999999999
Answers
11
+ 192 (Keep adding 9 between the digits ofnumbers 12 to 19)
+ 1993
+ 19994
+ 199995
+ 1999996
+ 19999997
+ 199999998
+ 1999999999
=2,22,22,22,175
This is the right answer.
I hope my answer will help u.....
Given : 11+192+1993+19994+199995......1999999999
To find : Sum
Solution:
11+192+1993+19994+199995......1999999999
There are total 10 numbers
Unit Digit is from 1 to 9
hence sum = 45
5 is the unit digit
4 is carried over
tens digits - 1 + 8 * 9 + 4 ( carried over) = 77
7 is tens digit
7 is carried over
Hundreds digit 1 + 7*9 + 7 (carried over) = 71
1 is hundreds digit
7 is carried over
Next = 1 + 6*9 + 7 = 62
2 digit
6 carried over
1 + 5*9 + 6 = 52
2 Digit
5 carried over
1 + 4*9 + 5 = 42
2 digit
4 carried over
1 + 3*9 + 4 = 32
2 Digit
3 carried over
1 + 2 * 9 + 3 = 22
2 Digit
2 Carried over
1 + 1 * 9 + 2 = 12
2 Digit
1 Carried over
1 + 1 = 2
2222222175 is the sum
11 + 192 + 1993 + 19994 + 199995 + 1999996 + 19999997 + 199999998 + 1999999999 = 2222222175
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