Question 2 (i)
Find the sum of those integers between 1 and 500 which are multiple of 2 as well as of 5.

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Solution

Mulitples of 2 as well as of 5 between 1 and 500 are 10, 20, 30 …. 490
nth term $a_{n}$ = last term (l) = 490
$∴$ Sum of n terms between 1 and 500,
$S_{n} = \frac{n}{2} [a + l]$ ...(i)
$∵ a_{n} = a + (n - 1) d = l$
$\Rightarrow 10 + (n - 1) 10 = 490$
$\Rightarrow (n - 1) 10 = 480$
$\Rightarrow n - 1 = 48 \Rightarrow n = 49$
$F r o m E q . (i) . . . S_{49} = \frac{49}{2} (10 + 490)$
$= \frac{49}{2} \times 500$
$= 49 \times 250 = 12250$

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Q. Find the sum of those integers between 1 and 500 which are multiples of 2 as well as of 5.

Q. Question 2 (ii)
Find the sum of those integers from 1 to 500 which are multiples of 2 as well as of 5.

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(i) all integers between $100$ and $550$ , which are divisible by $9$ .
(ii) all integers between $100$ and $550$ which are not divisible by $9$ .
(iii) all integers between $1$ and $500$ which are multiples of $2$ as well as of $5$ .
(iv) all integers from $1$ to $500$ which are multiplies $2$ as well as of $5$ .
(v) all integers from $1$ to $500$ which are multiples of $2$ or $5$ .

Q. Question 2 (iii)
Find the sum of those integers from 1 to 500 which are multiples of 2 or 5.

Q. Find the sum of those integers from

1

500

which are multiples of

2

5.

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Question 2 (i) Find the sum of those integers between 1 and 500 which are multiple of 2 as well as of 5.

Mulitples of 2 as well as of 5 between 1 and 500 are 10, 20, 30 …. 490 nth term an = last term (l) = 490 ∴ Sum of n terms between 1 and 500, Sn=n2[a+l] ...(i) ∵ an=a+(n−1)d=l ⇒10+(n−1)10=490 ⇒(n−1)10=480 ⇒n−1=48⇒n=49 From Eq. (i)... S49=492(10+490) =492×500 =49×250=12250

Question 2 (i)
Find the sum of those integers between 1 and 500 which are multiple of 2 as well as of 5.