Question

find the sum of all positive integers less than 200 that are divisible by 3 and 5 b) divisible by 3 not 5​

Answer 1

we will use the sum of natural number series formula

if S=1+2+3+4+........+n, then

S= n(n+1) ;

2

Now, sum of all +ve integers from 0 to 200 is given by

S=0+1+2+3+.......+200

=> S=200(200+1)/2;

=>S=20100;

Sum of all +ve integers divisible by 5 is given by

R=0+5+10+15+20+25+30+.........+200

R=5(1+2+3+4+5+6+.........+40)

R=5*40(40+1)/2

R=4100 ;

therefore the sum of all +ve integers <200 and not divisible by 5=S-R=16000

Answer 2

Answer:

(a)1365 (b) 5268

Step-by-step explanation:

(a)divisible by 3 and 5

Sol: let S be the required sum.

ATQ,

S = 15+30+45+60+............+195.

; a=15 & d= 15

Tn = 195

a+(n-1)d=195

15+(n-1)15=195

[ n = 13 ]

S = n/2 × {2a+(n-1)d}

S = 13/2 × {30+(12)15}

S = 13×105 = 1365 ans.

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(b) divisible by 3 but not by 5

Sol: let S be the required sum

ATQ,

S=sum of the nos. divisible by 3 -

sum of the nos. divisible

by 15(think u will get the reason)

. Let S1 be the sum of the nos.div.by 3.

Let S2 be the sum if nos. div.by 15.

(Remember we had already calculated S2 in previous part.)

S = S1 - S2

S = 6633-1365 = 5268 ans.

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