Find the sum of three digit numbers that are not divisible by 7 ?
Answers
Answer:
100 + 101 + 102 +...... + 999 [ The series is in A.P.] Let 994 be the nth term of the given series.
Answer:
70336
Step-by-step explanation:
Sumof3digitnumbers
divisibleby7=70336
Step-by-step explanation:
100,101,102,...,999 are 3 digit numbers.
105, 112, 119,....,994 are 3 digit numbers divisible by 7.
first term (a) = 105,
Last term (l) = 994
\begin{lgathered}common\: difference (d)\\=a_{2}-a_{1}\\=112-105\\=7\end{lgathered}
commondifference(d)
=a
2
−a
1
=112−105
=7
l = 994l=994
\implies a+(n-1)d = 994⟹a+(n−1)d=994
\implies 105+(n-1)7= 994⟹105+(n−1)7=994
/* Divide each term by 7, we get
\implies 15+n-1 = 142⟹15+n−1=142
\implies14+n=142⟹14+n=142
\implies n = 142 - 14⟹n=142−14
\implies n = 128⟹n=128
Sum \:of \:n \:terms(S_{n})=\frac{n}{2}(a+l)Sumofnterms(S
n
)=
2
n
(a+l)
\implies S_{128}=\frac{128}{2}(105+994)⟹S
128
=
2
128
(105+994)
= 64\times 1099=64×1099
=70336=70336
Therefore,
\begin{lgathered}Sum \:of \:3\:digit \: numbers\\divisible \:by \:7 = 70336\end{lgathered}
Sumof3digitnumbers
divisibleby7=70336