Question

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 $(a_1 + a_2 + …. + a_n ) > 1830?$

• A. 8
• B. 9
• C. 10
• D. 11

Answer 1

Answer: (B)

9

Answer 2

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, $m=9.$