Question

Find the sum of the integers between $100$ and $200$ that are divisible by $9$.

• A. $1659$
• B. $1693$
• C. $1653$
• D. $1683$

Answer 1

Answer: (D)

$1683$

the series is given by,

$108,117,\mathrm{126....}198$ and so on

So $a=108$

$l=198$

$n=?$

nth term$=a+(n-1)d$

$198=108+(n-1)9$

$198=108+9n-9$

$198-99=9n$

$n=11$

Now, sum of n terms

Sum of first 11 terms

$=\frac{11}{2}(108+198)$

$=\frac{11}{2}\left(306\right)$

$=11\times 153$

$=1683$