find the sumof all three digit number which is divisible by 13
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According to question the three-digit numbers which are divisible by 13 are 104, 117, 130, 143,.…… 938.This forms\xa0an AP in which a = 104, d = (117 – 104) = 13 and\xa0l = 938 (last term)Let the number of terms be n\xa0Then Tn\xa0= 938{tex}\\Rightarrow{/tex}a+(n-1)d=988{tex}\\Rightarrow{/tex}104+(n-1){tex}\\times{/tex}13=988{tex}\\Rightarrow{/tex}13n=897{tex}\\Rightarrow{/tex}n=69Therefore required sum={tex}\\frac{n}{2}{/tex}(a+l)={tex}\\frac{{69}}{2}{/tex}[104+988]=69{tex}\\times{/tex}546=37674Hence, the sum of all three digit\xa0numbers which are\xa0divisible by 13 is equal to 37674.