Consider an A.P. of positive integers,whose s | vedclass

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(C) Let the $A.P.$ be $a, a+d, a+2d, \dots$ where $a$ and $d$ are positive integers.Given $S_3 = a + (a+d) + (a+2d) = 3a + 3d = 54$,which simplifies t

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$90$

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(C) Let the $A.P.$ be $a, a+d, a+2d, \dots$ where $a$ and $d$ are positive integers.Given $S_3 = a + (a+d) + (a+2d) = 3a + 3d = 54$,which simplifies to $a+d = 18$.Thus,$a = 18-d$.Since $a$ is a positive integer,$18-d > 0 \Rightarrow d Also,$S_{20} = \frac{20}{2} [2a + 19d] = 10[2(18-d) + 19d] = 10[36 - 2d + 19d] = 10[36 + 17d]$.Given $1600 Subtracting $36$ gives $124 Dividing by $17$ gives $7.29 Since $d$ must be an integer,$d = 8$.Then $a = 18 - 8 = 10$.The $11^{	ext{th}}$ term is $a_{11} = a + 10d = 10 + 10(8) = 10 + 80 = 90$.

Consider an $A.P.$ of positive integers,whose sum of the first three terms is $54$ and the sum of the first twenty terms lies between $1600$ and $1800$. Then its $11^{\text{th}}$ term is:

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