Question

Find the sum of all the even positive and integers less than 200 which are not divisible by 6.
A) 6536
B) 6539
C) 6534
D) 6532

Answer 1

Hint: So, we will calculate the sum of all the even positive integers less than 200 first. Then we will find out how to sum all the products of 6 that 'fit into' 200. By subtracting these two we will be able to find the sum of all the even positive integers less than 200 which are not divisible by 6.
 The sum of all the even positive integers less than 200 which are not divisible by 6= (Sum 1 to 200) - (sum all the products of 6 that 'fit into' 200).

Complete step-by-step answer:
We will use the sum of natural number series formula
if S=1+2+3+4+........+n, then
\[S = \dfrac{{{\text{ }}n\left( {n + 1} \right)}}{2}\]
Now, sum of all even positive integers less than 200 is given by
\[\begin{array}{*{20}{l}}
S = 2 + 4 + 6 + 8 + 10 + 12....... + 198 \\
\Rightarrow S = 2(1 + 2 + 3 + 4 + 5 + 6....... + 99) \\
\\
{ \Rightarrow {\text{ }}S = 2\left( {\dfrac{{99\left( {99 + 1} \right)}}{2}} \right)} \\
{ = > S = 9900}
\end{array}\]
Sum of all positive integers less than 200 divisible by 6 is given by
\[\begin{array}{*{20}{l}}
{ \Rightarrow R = 0 + 6 + 12 + 18 + 24 + 30 + 36 + ......... + 198} \\
{ \Rightarrow R = 6\left( {1 + 2 + 3 + 4 + 5 + 6 + ......... + 33} \right)} \\
{ \Rightarrow R = 6 \times \dfrac{{33\left( {33 + 1} \right)}}{2}} \\
{ \Rightarrow R = 3366{\text{ }}}
\end{array}\]
Therefore, the sum of all even positive integers less than 200 and not divisible by 6 \[ = S - R = 9900 - 3366 = 6534\]

So, option (C) is the correct answer.

Note: A geometric sequence goes from one term to the next by always multiplying (or dividing) by the same value. So 1, 2, 4, 8, 16,... is geometric, because each step multiplies by two; and 81, 27, 9, 3, 1,... is geometric, because each step is divided by 3. The number multiplied (or divided) at each stage of a geometric sequence is called the "common ratio" r, because if you divide (that is, if you find the ratio of) successive terms, you'll always get this common value.
So, in such questions before finding the sum you take the common ratio common to make the calculation easier.