Find the sum of the ontegers between 100 and 200 thar are divisible by 9
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First no a=108
last no l=198
d=9
total no tn=a+(n-1)d
198=108+9n-9
9n=198-99
n=99/9
n=11
sum of 11 no is
sn=n/2(a+l)
=11/2(108+198)
=11/2×306
=11×153
sn=1683
last no l=198
d=9
total no tn=a+(n-1)d
198=108+9n-9
9n=198-99
n=99/9
n=11
sum of 11 no is
sn=n/2(a+l)
=11/2(108+198)
=11/2×306
=11×153
sn=1683
Answered by
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First term = a = 108
Last term = l = 198
=> 198 = 108 + ( n - 1 ) 9
=> 198 - 108 = ( n - 1 ) 9
=> 90 = ( n - 1 ) 9
=> 10 = n - 1
=> 11 = n
Hence, Number of term = n = 11
So,
Last term = l = 198
=> 198 = 108 + ( n - 1 ) 9
=> 198 - 108 = ( n - 1 ) 9
=> 90 = ( n - 1 ) 9
=> 10 = n - 1
=> 11 = n
Hence, Number of term = n = 11
So,
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