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Find the sum of all natural numbers not exceeding $1000,$ which are divisible by $4$ but not by $8 .$

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62500

62800

64000

65600

answer is A.

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Detailed Solution

In the given statement the natural numbers below 1000, that are divisible by 4 but not by 8 are as below,

4, 12, 20, 28, . . ., 996

These numbers form an AP with

a = 4, d = 8

and,

a_{n} = 996

.
We know that for an AP,

a_{n} = a + (n - 1) d

\Rightarrow 996 = 4 + (n - 1) \times 8

\Rightarrow 8 n - 8 + 4 = 996

\Rightarrow n = 125

For sum of n terms in an AP,

S_{n} = \frac{n}{2} \times [2 a + (n - 1) d]

⇒

S_{125} = \frac{125}{2} [2 \times 4 + (125 - 1) \times 8]

\Rightarrow \frac{125}{2} \times [8 + 992]

\Rightarrow \frac{125}{2} \times 1000

\Rightarrow 62500

Hence, the correct option is 1.

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