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CAT 2017 [slot 1] Question with solution 34

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?
  1. 8
  2. 9
  3. 10
  4. 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.


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