Find the sum of numbers divisible by 123 in between 1000 and 2000.
Answers
Step-by-step explanation:
Numbers divisible by 123 have clearly a difference of 123 and thus forms an A.P.
First number divisible by 123 between 1000 and 2000 is 1107.
Last number divisible by 123 between 1000 and 2000 is 1968 .
Thus ,
a = 1107
l = 1968
d = 123
Now , let there be 'n' numbers of terms , then ;
=) a + ( n - 1 ) d = 1968
=) 1107 + ( n - 1 )* 123 = 1968
=) 1107 + 123n - 123 = 1968
=) 123n = 1968 - 1107 + 123
=) 123n = 984
=) n = ( 984 / 123 ) = 8
Hence , total number of terms is 8 .
Now , by the sum formula , we have
Sum = ( n / 2 ) * { 2a + ( n - 1 ) d }
= ( 8 / 2 ) * { ( 2 * 1107 ) + ( 7 * 123 ) }
= 4 ( 2214 + 861 )
= 4 ( 3075 )
= 12300
Hence , the sum of numbers divisible by 123 between 1000 and 2000 is 12300 .
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Numbers divisible by 123 in between 1000 and 2000.
a = 1107
d = 123
L = 1968
L = a + (n-1)d
=> 1986 - 1107 = (n-1) 123
=> 861 / 123 = (n-1)
=> n-1 = 7
=> n = 7+1
=> n = 8
S = n/2 ( 2a + (n-1)d)
=> S = 8/2 (1107+1986)
=> S = 4 × 3075
=> S = 12,300 Ans.
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