12. Find the sum of all numbers between 100 and 200 which are not divisible by 5.
Answers
101,102,103,.....,199 ......(i)
and
105,110,115,.....,195 ......(ii)
in (i),
a=101, d=1 and tn=199
tn=a+(n-1)d
199=101+(n-1)1
199=101+n-1
199=100+n
n=99
Sn=n/2 {2a+(n-1)d}
=99/2 {2*101 +(99-1)1}
=99/2 {202+98}
=99/2 {300}
=99*150=14850
in (ii)
a=105, d=5 and tn=195
tn=a+(n-1)d
195=105+(n-1)5
(n-1)5=90
n-1=18
n=19
Sn=n/2 {2a+(n-1)d}
=19/2 {2*105 +(19-1)5}
=19/2 {210+90}
=19/2 {300}
=19*150=2850
therefore the sum of all number between 100 and 200 which are not divisible by 5=
the sum of all number between 100 and 200 - the sum of all number between 100 and 200 which are divisible by 5
=14850-2850
=12000
HOPE IT HELPS YOU
i ) Numbers between 100 and 200
are 101 , 102 , ...., 199 are in A.P
First term = a = a1 = 101 ,
Common difference = d
d = a2 - a1
d = 102 - 101 = 1
nth term = an
an = 199
a + ( n - 1 )d = 199
=> 101 + ( n - 1 )1 = 199
=> 100 + n = 199
n = 99
ii ) Sum n terms = Sn
Sn = n/2 [ a + an ]
=> S99= 99/2 [ 101 + 199 ]
= 99 × 150
= 14850 ---( 1 )
iii ) Numbers between 100 and 200
which are divisible by 5 are ,
105 , 110, 115 , ...., 195 are in A.P
First term = a = a1 = 105
Common difference = d
d = a2 - a1
d = 110 - 105 = 5
nth term = 195
a + ( n - 1 )d = 195
=> 105 + ( n - 1 )5 = 195
=> ( n - 1 )5 = 195 - 105
=> ( n - 1 )5 = 90
=> n - 1 = 90/5
=> n = 18 + 1
n = 19
ii ) Sum n terms = Sn
Sn = n/2[ a + an ]
S19 = 19/2 [ 105 + 195 ]
S19 = 19/2 × 300
S19 = 2850
Sum of numbers which are
devisible by 5 are
S19 = 2850----( 2 )
iv ) Sum of all numbers between
100 and 200 which are not
divisible by 5 = ( 1 ) - ( 2 )
= 14850 - 2850
= 12000
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