Find the sum of the integers between 100 and 200 that are (i) divisible by 9 (ii) not divisible by 9
Answers
Answer:
i) 1683
ii)13367
Step-by-step explanation:
i) Numbers between 100 to 200 that are divisible by 9 ,form an A.P.
A.P.: 108 , 117 , .........., 198
here, a = 108 , an = 198 ,
d = a2-a1 = 117-108 = 9
We know that,
an = a +(n-1)×d
By applying the values we get,
198 = 108 + (n-1) × 9
198 = 108 + 9n - 9
198 = 99 + 9n
9n = 198 - 99
9n = 99
n = 11
Hence, number of terms between 100 to 200 that are divisible
by 9 = n = 11
Now,
Sn = n(a+an)/2
By applying values,we get
S11 = 11(108+198) / 2
S11 = ( 11 × 306 ) / 2
S11 = 11 × 153
S11 = 1683
Hence the sum of integers that are divisible by 9 between 100 and 200 is 1683.
ii) New A.P. of numbers between 100 to 200 =>
A.P.: 101 , 102, 103,......................, 200
Here, a = 101 , an = 200
d = a2- a1 = 102-101 = 1
We know that,
an = a + (n-1)× d
200 = 101 + (n-1) × 1
200 = 101 + n - 1
200 = 100 + n
n = 200-100 = 100
Hence numbers of terms between 100 and 200 = n = 100
Now,
Sn = n(a+an) / 2
By applying values,we get
S100 = 100(101+200) / 2
S100 = 50 × 301
S100 = 15050
Now, sum of integers between 100 to 200 that are not divisible
by 9 = sum of integers between 100 to 200 - sum of integers between 100 to 200 that are divisible by 9
= S100 - S11
= 15050 - 1683
= 13367
Hence sum of integers between 100 and 200 that are not divisible by 9 = 13367