Question

Obtain the sum of all positive integers up to 1000, which are divisible by 5 and not divisible by 2
(a)10050
(b)5050
(c)5000
(d)50000

Answer 1

Hint: For solving this problem, we have to first find out the numbers which are divisible by 5 up to 1000. Then, we have to eliminate the numbers which are divisible by 2. Now, we evaluate the number of terms in this arithmetic progression. Finally, we obtain the sum of series by using the formula of A.P.

Complete step-by-step answer:
According to our problem, the positive integers which are divisible by 5 are 5,10,15,20........1000 now, the integers out of these which are divisible by 2 are 10,20,30..............1000. Eliminating them, we get the final series as 5,15,25.....995.
Thus, we have to find the sum of positive integers 5,15,25......995
If n is the number of terms in it, so by using the general term for an A.P.: $a_n=a+(n-1)\cdot d$
Now, putting values of variables as $a_n$ = 995, a = 5 and d = 10, we get
$\begin{array}{rl}& 995=5+(n-1)\cdot 10\\ & 995-5=(n-1)\cdot 10\\ & 990=(n-1)\cdot 10\\ & n-1={\displaystyle \frac{990}{10}}\\ & n-1=99\\ & n=100\end{array}$
Thus, the sum of numbers in A.P.: $S_n={\displaystyle \frac{n}{2}}(a+l)$
So, putting values of variables as n = 100, a = 5 and l = 995, we get
$\begin{array}{rl}& S={\displaystyle \frac{100}{2}}(5+995)\\ & S=50\times 1000\\ & S=50000\end{array}$
Therefore, the sum of all positive integers up to 1000, which are divisible by 5 and not divisible by 2 is 50000.
Hence, option (d) is correct.
Note: This problem can alternatively be solved by using the general formula for sum of series in arithmetic progression which can be stated as $S={\displaystyle \frac{n}{2}}(2a+(n-1)\cdot d)$. One must remember that for solving any problem in which a number of terms are required, we can form an A.P. and then apply the formula to evaluate the number of terms.