6. Find the sum of whole number divisible by 3 and lies between 100 and 200.

4950

Step-by-step explanation:

first find the smallest multiple of 3 thats >100, 3*33=99, so 3*34=102 is the smallest multiple of 3 >100.

next find the greatest multiple of 3 <200, 3*66=198 which is the greatest multiple of 3<200

34-66+1=number of terms=33. then find the average of the terms: 102+198/2=150. 150*33=4950. the answer should be 4950

Step-by-step explanation:

The first number in the sequence = 102

The last number = 198

The number of terms =

L = a1 + (n-1)*d

198 = 102 + (n- 1)*3                   Subtract 102

198 - 102 = (n - 1)* 3

96 = (n - 1)*3                             divide by 3

32 = n - 1                                   add 1

33 = n

Sum = (a1 + L ) * n/2

sum = (102+198)*33/2

sum = 300 * 33/2

sum = 150 * 33

sum = 4950

hello from mrbilldoesmath!

answer:

a(n) - a(n-1) -3

discussion:

this ones a bit interesting. given a(n) = 5 - 3n consider this:

a(n-1) = 5 - 3(n-1)

          = (5 - 3n) + 3             => but a(n) = 5 - 3n, the first term

          = a(n)   + 3                 => subtract 3 from both sides

a(n-1) -3 = a(n)  

or

a(n) - a(n-1) -3

you,

mrb

add 5 subtract 1

step-by-step explanation: