Question 34: 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?
8
9
10
11
Option: 2 Explanation:
$a_{1}$ = 3 and $a_{2}$ = 7. Hence, the common difference of the AP is 4. We have been given that the sum up to 3n terms of this AP is 1830. Hence, $1830 = \frac{m}{2}[2×3 + (m - 1)×4$ => 1830×2 = m(6 + 4m - 4) => 3660 = 2m + 4m$^2$ => $2m^2 + m - 1830 = 0$ => (m - 30)(2m + 61) = 0 => m = 30 or m = -61/2 Since m is the number of terms so m cannot be negative. Hence, must be 30 So, 3n = 30 n = 10 Sum of the first '10' terms of the given AP = 5×(6 + 9×4) = 42×5 = 210 m($a_1$+ $a_{2}$ +...+ $a_n$) > 1830 => 210m > 1830 => m > 8.71 Hence, smallest integral value of 'm' is 9.