Question

find the sum of the integers between 100 and 200 1) divisible by 3. 2) not divisible by 3​

Answer 1

Answer:

101 + 102 + 103.............

Answer 2

Step-by-step explanation:

1) divisible by 3

first number (a) = 102

common difference (d) = 3

to find the number if terms in this AP :

n: no. of terms

l = 198

last term (l) = a + (n-1)d

198 = 102 + (n-1) 3

96 = (n-1) 3

32 = n-1

=> n= 33

Sum of n terms = n/2( 2a + (n-1)d)

S = 33/2( 2*102 + (33-1)*3)

S = 33/2( 204 + 32*3)

S = 33/2( 204 + 96)

S = 33/2 ( 300)

S= 33*150

S = 4950 : Answer 1.

2) not divisible by 3

we'll use the sum we found in part 1 by subtracting it from sum of all terms between 100 and 200.

101, 102, 103.....199

Let S1 be the sum of above series and n be the no. of terms.

for n:

l = a+ (n-1) d

199 = 101 + (n-1) 1

98 = n-1

n= 99

S1 = 99/2( 2*101 + (99-1) 1)

S1 = 99/2( 202 + 98)

S1 = 99/2 ( 300)

S1 = 99 * 150

S1 = 14850

Sum of terms not divisible by 3

= S1 - S

= 14850 - 4950

= 9900. : answer

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