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CAT 2017 Slot 1 QA Question & Solution

AlgebraMedium

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?

Options

8
9
10
11

Solution

$a_{1}$ = 3 and $a_{2}$ = 7. Hence, the common difference of the AP is 4. If we assume, k=3n 
We have been given that the sum up to 3n terms of this AP is 1830. Hence, $1830 = \frac{k}{2}[2*3 + (k - 1)*4$
=> 1830*2 = k(6 + 4k - 4)
=> 3660 = 2k + 4k$^2$
=> $2k^2 + k - 1830 = 0$
=> (k - 30)(2k + 61) = 0
=> k = 30 or k = -61/2
Since k is the number of terms so k 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.