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| 1. |
Find the sum of all the three digit numbers which leave remainder 2 when divided by 5 |
| Answer» Three digit numbers which leave the remainder 2 when divided by 5 are 102, 107, 112, 117......, 997.102, 107, 112.... 997 is an A.PIn this A.Pa (first term)= 102d (Common difference)= 5I(last term ) = 997l= an = a + (n – 1) d997= 102 + (n – 1) {tex}\\times{/tex} 55 (n – 1) = 997 – 102 = 895n-1={tex}\\frac {885}{5}{/tex}(n – 1) = 179n = 179 +1n = 180Sum of all three digit numbers which leaves remainder 2 when divided by 5Sn ={tex}\\frac{n}{2}{/tex} [a+l]Sn= {tex}\\frac{180}{2}{/tex} [ 102+ 997]Sn= 90 {tex}\\times{/tex} 1099Sn= 98910Sum of all three digit numbers which leaves remainder 2 when divided by 5 is 98910\xa0 | |