Question

Find the sum of all odd numbers of four digits which are divisible by $9.$

Answer 1

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