Find the sum of those integers from 1 to 500 | vedclass

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(B) To find the sum of integers from $1$ to $500$ which are multiples of $2$ or $5$, we use the principle of inclusion-exclusion:Sum $= (	ext{Sum of

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$75250$

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(B) To find the sum of integers from $1$ to $500$ which are multiples of $2$ or $5$, we use the principle of inclusion-exclusion:Sum $= (	ext{Sum of multiples of } 2) + (	ext{Sum of multiples of } 5) - (	ext{Sum of multiples of } 	ext{LCM}(2, 5) = 10)$.$1$. Multiples of $2$: $2, 4, 6, \dots, 500$. This is an $AP$ with $a=2, l=500, d=2$.Number of terms $n_1$: $500 = 2 + (n_1 - 1)2 \Rightarrow 498 = 2(n_1 - 1) \Rightarrow n_1 - 1 = 249 \Rightarrow n_1 = 250$.Sum $S_1 = \frac{250}{2}(2 + 500) = 125 	imes 502 = 62750$.$2$. Multiples of $5$: $5, 10, 15, \dots, 500$. This is an $AP$ with $a=5, l=500, d=5$.Number of terms $n_2$: $500 = 5 + (n_2 - 1)5 \Rightarrow 495 = 5(n_2 - 1) \Rightarrow n_2 - 1 = 99 \Rightarrow n_2 = 100$.Sum $S_2 = \frac{100}{2}(5 + 500) = 50 	imes 505 = 25250$.$3$. Multiples of $10$: $10, 20, 30, \dots, 500$. This is an $AP$ with $a=10, l=500, d=10$.Number of terms $n_3$: $500 = 10 + (n_3 - 1)10 \Rightarrow 490 = 10(n_3 - 1) \Rightarrow n_3 - 1 = 49 \Rightarrow n_3 = 50$.Sum $S_3 = \frac{50}{2}(10 + 500) = 25 	imes 510 = 12750$.Total Sum $= S_1 + S_2 - S_3 = 62750 + 25250 - 12750 = 88000 - 12750 = 75250$.

Find the sum of those integers from $1$ to $500$ which are multiples of $2$ or $5$.

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