Question

find out the sum of all natural numbers between 1 and 145 which are divisible by 4 arithmatic progression

Answer 1

Answer:

264

Step-by-step explanation:

do not include 1 and 45 in this sum.

n=  number of terms divisible by 4(i.e,11)

n=43

a= first number divisible by four=4

l=last number divisible by 4=44

so sum of the 11 terms=sum of all natural numbers b/w 1 and 45 divisible by 4.

S11= 11(4+44)/2

S11= 11(48)/2

S11= 11(24)

S11= 264

Answer 2

Answer:

answer is 2664.

Step-by-step explanation:

a=4, d=4, tn=144

tn=a+(n-1)*d.....(formula)

144=4+(n-1)*4

144=4+4n-4

144=4n

144/4=n

n=36

   sum of all natural numbers between 1 to 145 which are divisible by 4 is

Sn=n/2(t1+tn).........(formula)

S36=36/2(4+144)

S36=18*148

S36=2664

  So,the sum of all natural numbers between 1 and 145 which are divisible by 4 is 2664.