The series of positive multiples of 3 is divided into sets : $\left\{3\right\},\left\{6,9,12\right\},\left\{15,18,21,24,27\right\},......$ Then the sum of the elements in the 11th set is equal to…………..

<p>Given series&nbsp;<math><mrow><mrow><mo> {</mo><mrow><mn>3</mn><mo>×</mo><mn>1</mn></mrow><mo>}</mo></mrow><mo>, </mo><mrow><mo> {</mo><mrow><mn>3</mn><mo>×</mo><mn>2</mn><mo>, </mo><mn>3</mn><mo>×</mo><mn>3</mn><mo>, </mo><mn>3</mn><mo>×</mo><mn>4</mn></mrow><mo>}</mo></mrow><mo>, </mo><mrow><mo> {</mo><mrow><mn>3</mn><mo>×</mo><mn>5</mn><mo>, </mo><mn>3</mn><mo>×</mo><mn>6</mn><mo>, </mo><mn>3</mn><mo>×</mo><mn>7</mn><mo>, </mo><mn>3</mn><mo>×</mo><mn>8</mn><mo>, </mo><mn>3</mn><mo>×</mo><mn>9</mn></mrow><mo>}</mo></mrow><mo>.</mo><mo>.</mo><mo>.</mo><mo>.</mo><mo>.</mo><mo>.</mo><mo>.</mo><mo>.</mo><mo>.</mo></mrow></math></p><p><math><mrow><mo>∴</mo></mrow></math>&nbsp;11th set will have 1 + (10)2 = 21 terms</p><p>Also up to 10th set total 3 × k type terms will be 1 + 3 + 5 + ……… +19 = 100 terms</p><p><math><mrow><mo>∴</mo><mi>S</mi><mi>e</mi><mi>t</mi><mtext> </mtext><mtext> </mtext><mn>1</mn><mn>1</mn><mo>=</mo><mrow><mo> {</mo><mrow><mn>3</mn><mo>×</mo><mn>1</mn><mn>0</mn><mn>1</mn><mo>, </mo><mn>3</mn><mo>×</mo><mn>1</mn><mn>0</mn><mn>2</mn><mo>, </mo><mo>.</mo><mo>.</mo><mo>.</mo><mo>.</mo><mo>.</mo><mo>.</mo><mn>3</mn><mo>×</mo><mn>1</mn><mn>2</mn><mn>1</mn></mrow><mo>}</mo></mrow></mrow></math>&nbsp;<math><mrow><mo>∴</mo></mrow></math>&nbsp;Sum of elements = 3 × (101 + 102 + ….+121)</p><p><math><mrow><mo>=</mo><mfrac><mrow><mn>3</mn><mo>×</mo><mn>2</mn><mn>2</mn><mn>2</mn><mo>×</mo><mn>2</mn><mn>1</mn></mrow><mrow><mn>2</mn></mrow></mfrac><mo>=</mo><mn>6</mn><mn>9</mn><mn>9</mn><mn>3</mn><mo>.</mo></mrow></math></p>

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