Sequences and Series

The integers divisible by 5 from 1 to 100 are $5,10,15,20,25,\ldots,100$. Which are in A.P. with first term (a) = 5 and common difference $(d)=5$ ) last term $(l)=100$So, total number of terms,$(n)\;=\;\frac{l-a}{d}+1\;=\;\frac{100-5}{5}+1\;=\;\frac{95}{5}+1$$\;=19+1=20$$\therefore S_n\;=\;\frac{n}{2}(a+l)$$\;=10(5+100)=1050$Similarly, the integers divisible by 13 from 1 to 100 are $13,26,39,52,\ldots,91$ which are in AP with to $a=13, d=13, l=91$.$\;S o, n\;=\;\frac{l-a}{d}+1\;=\;\frac{91-13}{13}+1\;=\;\frac{78}{13}+1$$\;=6+1=7$$\;\therefore \;\; S_n\;=\;\frac{7}{2}(13+91)=7 \times 52=364$Now, the integer divisible by both 5 and 13 from 1 to 100 is 65 .Hence, the sum of integer from 1 to 100 that are divisible by 5 or 13 is$=1050+364-65=1414-65=1349$