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Find the sum of all natural numbers lying between 100 and 1000, which are multiple of 5

Posted by Junaid Saifi 4 years, 9 months ago

CBSE > Class 11 > Mathematics

1 answers

Sia ? 4 years, 9 months ago

According to question, series is 105, 110, 115, 120, ………, 995
Here a = 105, d = 110 - 105 = 5 and a_n = 995
{tex}\therefore{/tex} a_n = a + (n - 1)d
{tex}\Rightarrow 995 = 105 + ( n - 1 ) \times 5{/tex}

{tex}\Rightarrow 995 - 105 = ( n - 1 ) \times 5{/tex}

{tex}\Rightarrow \frac { 890 } { 5 } = ( n - 1 ){/tex}

{tex}\Rightarrow{/tex} n = 178 + 1 = 179

Now, {tex}\mathrm { S } _ { n } = \frac { n } { 2 } ( a + l ){/tex}

{tex}\Rightarrow S _ { 199 } = \frac { 179 } { 2 } ( 105 + 995 ){/tex}

{tex}\Rightarrow S _ { 179 } = \frac { 179 } { 2 } \times 1100=98450{/tex}

1Thank You

ANSWER

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CBSE > Class 11 > Mathematics