Question Number 57062 by olalekan2 last updated on 29/Mar/19
$${find}\:{the}\:{sum}\:{of}\:{all} \\ $$$${three}\:{digital}\:{natural} \\ $$$${numbers}\:{that}\:{are}\: \\ $$$${divisible}\:{by}\:\mathrm{7} \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 29/Mar/19
Commented by tanmay.chaudhury50@gmail.com last updated on 29/Mar/19
Answered by tanmay.chaudhury50@gmail.com last updated on 29/Mar/19
![first three digit number divisble by 7 is=105 second is=(105+7=112) ... ... last three digit number divible by 7=994 so S=105+112+...+994 in A.p series a=105 d=7 a+(n−1)7=994 105+(n−1)7=994 n−1=((994−105)/7)=((889)/7)=127 n=128 S=(n/2)[2a+(n−1)d] =((128)/2)[a+a+(n−1)d] =((128)/2)[105+994]=70336](Q57070.png)
$${first}\:{three}\:{digit}\:{number}\:{divisble}\:{by}\:\mathrm{7}\:{is}=\mathrm{105} \\ $$$${second}\:{is}=\left(\mathrm{105}+\mathrm{7}=\mathrm{112}\right) \\ $$$$... \\ $$$$... \\ $$$${last}\:{three}\:{digit}\:{number}\:{divible}\:{by}\:\mathrm{7}=\mathrm{994} \\ $$$${so}\:{S}=\mathrm{105}+\mathrm{112}+...+\mathrm{994} \\ $$$$ \\ $$$${in}\:{A}.{p}\:{series}\:{a}=\mathrm{105}\:\:\:{d}=\mathrm{7} \\ $$$${a}+\left({n}−\mathrm{1}\right)\mathrm{7}=\mathrm{994} \\ $$$$\mathrm{105}+\left({n}−\mathrm{1}\right)\mathrm{7}=\mathrm{994} \\ $$$${n}−\mathrm{1}=\frac{\mathrm{994}−\mathrm{105}}{\mathrm{7}}=\frac{\mathrm{889}}{\mathrm{7}}=\mathrm{127} \\ $$$${n}=\mathrm{128} \\ $$$${S}=\frac{{n}}{\mathrm{2}}\left[\mathrm{2}{a}+\left({n}−\mathrm{1}\right){d}\right] \\ $$$$=\frac{\mathrm{128}}{\mathrm{2}}\left[{a}+{a}+\left({n}−\mathrm{1}\right){d}\right] \\ $$$$=\frac{\mathrm{128}}{\mathrm{2}}\left[\mathrm{105}+\mathrm{994}\right]=\mathrm{70336} \\ $$
Answered by mr W last updated on 29/Mar/19
$${A}.{P}.: \\ $$$$\mathrm{105},\:\mathrm{112},\:\mathrm{119},\:...,\mathrm{994} \\ $$$${n}=\frac{\mathrm{994}−\mathrm{105}}{\mathrm{7}}+\mathrm{1}=\mathrm{128} \\ $$$$\Sigma=\frac{\left(\mathrm{105}+\mathrm{994}\right)×\mathrm{128}}{\mathrm{2}}=\mathrm{70336} \\ $$