Let \(a_1 , a_2 ,……..a_{3n}\) be an arithmetic progression with \(a_1 = 3\) and \(a_2 = 7.\) If \(a_1 + a_2 + ….+a_{3n} = 1830\), then what is the smallest positive integer m such that m \((a_1 + a_2 + …. + a_n ) > 1830?\)
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The Correct Option is B
Solution and Explanation
Given \(a_1=3\), it follows that \(d=a_2−a_1=7−3=4\) and \(a_2=7. \)
The sum \(a_1+a_2+a_3+…+a_{3n}=1830\).
Using the formula for the sum of an arithmetic series,
\(S_n=\frac{n}{2}[2a_1+(n−1)d],\)
we get \(1830=3n^2[2×3+(3n−1)×4]. \)
Simplifying this equation, we get \(12n^2+2n−610=0, \) which factors to \((6n+61)(n−10)=0. \)
Since n cannot be negative, \(n=10.\)
Substituting \(n=10\) into \(m(3+7+11+…+39)>1830, \)
we get \(m×10^2[2×3+(10−1)×4]>1830, \)
which simplifies to \(m×5(6+36)>1830. \)
Further simplification yields \(m×5×42>1830\), and solving for \(m\), we find \(m>8.714. \)
Therefore, \(m=9.\)
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