Prove the following

Solution:

(i) Since, multiples of 2 as well as of 5 = LCM of (2, 5) = 10

∴  Multiples of 2 as well as of 5 between 1 and 500 is 10, 20, 30. 490

which form an AP with first term (a) = 10 and common difference (d) = 20 -10 = 10

nth term an =Last term (/) = 490

∴ Sum of n terms between 1 and 500,

$S_{n}=\frac{n}{2}[a+l]$ ...(1)

$\because$ $a_{n}=a+(n-1) d=l$

$\Rightarrow \quad 10+(n-1) 10=490$

$\Rightarrow \quad(n-1) 10=480$

$\Rightarrow \quad n-1=48 \Rightarrow n=49$

From Eq. (i), $S_{49}=\frac{49}{2}(10+490)$

$=\frac{49}{2} \times 500$

 

$=49 \times 250=12250$

(ii) Same as part (i),

hut multinles of 2 as well as of 5 from 1 to 500 is $10,20,30, \ldots, 500$.

$\therefore$ $a=10, d=10, a_{n}=l=500$

$\because$ $a_{n}=a+(n-1) d=l$

$\Rightarrow \quad 500=10+(n-1) 10$

$\Rightarrow \quad 490=(n-1) 10$

$\Rightarrow \quad n-1=49 \Rightarrow n=50$

$S_{n}=\frac{n}{2}(a+l)$

$\Rightarrow \quad S_{50}=\frac{50}{2}(10+500)=\frac{50}{2} \times 510$

$=50 \times 255=12750$

(iii) Since, multiples of 2 or $5=$ Multiple of $2+$ Multiple of 5 - Multiple of $\mathrm{LCM}(2,5)$ i.e., 10 .

$\therefore$ Multiples of 2 or 5 from 1 to 500

$=$ List of multiple of 2 from 1 to $500+$ List of multiple of 5 from t to 500

- List of multiple of 10 from 1 to 500

$=(2,4,6, \ldots, 500)+(5,10,15, \ldots, 500)-(10,20, \ldots, 500)$ $\ldots$ (i)

All of these list form an AP.

Now, number of terms in first list,

$500=2+\left(n_{1}-1\right) 2 \Rightarrow 498=\left(n_{1}-1\right) 2$

$\Rightarrow$ $n_{1}-1=249 \Rightarrow n_{1}=250$

Now, number of terms in first list,

$500=2+\left(n_{1}-1\right) 2 \Rightarrow 498=\left(n_{1}-1\right) 2$

$\Rightarrow$$n_{1}-1=249 \Rightarrow n_{1}=250$  

Number of terms in second list,

$500=5+\left(n_{2}-1\right) 5 \Rightarrow 495=\left(n_{2}-1\right) 5$

$\Rightarrow \quad 99=\left(n_{2}-1\right) \Rightarrow n_{2}=100$

and number of terms in third list,

$500=5+\left(n_{2}-1\right) 5 \Rightarrow 495=\left(n_{2}-1\right) 5$

$\Rightarrow \quad n_{3}-1=49 \Rightarrow n_{3}=50$

From Eq. (i), Sum of multiples of 2 or 5 from 1 to 500

$=\operatorname{Sum}$ of $(2,4,6, \ldots, 500)+\operatorname{Sum}$ of $(5,10, \ldots, 500)-\operatorname{Sum}$ of $(10,20, \ldots, 500)$

$=\frac{n_{1}}{2}[2+500]+\frac{n_{2}}{2}[5+500]-\frac{n_{3}}{2}[10+500]$ $\left[\because S_{n}=\frac{n}{2}(a+l)\right]$

$=\frac{250}{2} \times 502+\frac{100}{2} \times 505-\frac{50}{2} \times 510$

$=250 \times 251+505 \times 50-25 \times 510=62750+25250-12750$

$=88000-12750=75250$