Question 100

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?

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.

Video Solution

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