Question

find the sum of all the three digit each of which leaves remainder3 when divisible by 5​

Answer 1

$\sf\large\underline{ Question :}$

Fund the sum of all the three digit each of which leaves remainder 3 when divisible by 5.

$\sf\large\underline{ Answer : }$

★ Given that,

Three digit numbers which are divisible by 5 and leaves remainder 3 are :

AP : 103, 108, 113, ..... 998.

★ To find,

• Sum of the AP series.

★ Let,

• $\bf\: a_{1} = 103$
• $\bf\: a_{2} = 108$

Common difference (d) = a2 - a1

$\bf\: \implies 108 - 103$

$\bf\: \implies 5$

Hence, the common difference (d) is 5.

★ Now,

• a = 103
• d = 5
• an = 998
• n = ?

By using nth formula we can get the value of n.

$\tt\purple{ \: a_{n} = a + (n - 1)d }</p><p>$

• Substitute the values.

$\bf\: \implies 998 = 103 + (n - 1)(5)$

$\bf\: \implies 998 = 103 + 5n - 5$

$\bf\: \implies 998 = 98 + 5n$

$\bf\: \implies 998 - 98 = 5n$

$\bf\: \implies 5n = 900$

$\bf\: \implies n = \frac{900}{5}$

$\bf\: \implies n = 180$

Hence, total no. of terms were 180.

★ Now we can find sum of 180 terms,

By using sum of n terms of AP, we can get value of Sn.

$\tt\purple{ S_{n} = \frac{n}{2} [ 2a + (n - 1)d ]}$

• Substitute the values.

$\bf\: \implies \frac{180}{2} [ 2(103) + (180 - 1)5 ]$

$\bf\: \implies 90 [ 206 + (179)5 ]$

$\bf\:\implies 90 [ 206 + 895 ]$

$\bf\:\implies 90 [ 1101 ]$

$\bf\:\implies 99090$

$\underline{\boxed{\bf{\red{ \therefore\:Hence,\:the\:sum\:of\:the\:given\:AP\:series\:is\:99090.}}}}\:\orange{\bigstar}$