Question

In the sequence of three digit numbers below 500 which leaves reminder 1 when divided by 7.vwrite the first term and the last term .

Answer 1

$a_{n} =l=498$⇒

$a_{n} =a_{1} +(n-1)d$⇒

$498=8+(n-1)7$⇒

$498=8+7n-7$⇒

$498-1=7n$⇒

$71=n$

$S_{n} =\frac{n}{2} (a+l)$⇒

$S_{71} =\frac{71}{2} (8+498)$⇒

$S_{71} =\frac{71}{2} (506)=17.963$

Step-by-step explanation: