Question

Find the sum of all odd numbers of four digits which are divisible by \[9.\]

Answer 1

Hint: Use the formula of the \[{{n}^{th}}\] term from beginning
\[{{T}_{n}}=a+(n-1)d\]
Where a = first term
d = common difference
n = number of the term
Use the formula of sum of \[n\] terms
\[{{S}_{n}}=\dfrac{n}{2}[2a+(n-1)d]\]

Complete step-by-step answer:
Four digit numbers, divisible by \[9\]are
\[1017,\,1035,..........9999\]
Sequence is in A.P with first term a = \[1017\]
Common difference = \[{{t}_{2}}-{{t}_{1}}\]
\[=\,1035-1017\]
\[=\,18\]
\[{{T}_{n}}=9999\]
Therefore
Use the formula of nth term from beginning is
\[{{T}_{n}}=a+(n-1)d\]
\[9999=1017+(n-1)18\]
\[9999-1017=18n-18\]
Simplify the expression
\[18n=8982+18\]
Rewrite the expression after simplification
\[18n=9000\]
\[n=\dfrac{9000}{18}\]
\[n=500\]
Use the formula of the sum of the nth term is
\[{{S}_{n}}=\dfrac{n}{2}[2a+(n-1)d]\]
Put the value of a, n and d
\[=\dfrac{500}{2}[2\times 1017+(500-1)\times 18]\]
Simplify the expression
\[=250[2034+499\times 18]\]
\[=250\times 11016\]
Rewrite the expression after simplification
\[=2754000\]

Required sum is equal to 2754000.

Note: This type of problem is also solved with the help of the formula.
\[{{S}_{n}}=\dfrac{n}{2}[{{a}_{1}}+{{a}_{n}}]\]
Where
\[{{a}_{1}}=\]First term
\[{{a}_{n}}=\]Last term