For any natural number n, suppose the sum of the first n terms of an arithmetic progression is $$(n + 2n^2)$$. If the $$n^{th}$$ term of the progression is divisible by 9, then the smallest possible value of n is
It is given,
$$S_n=2n^2+n$$
$$S_{n-1}=2\left(n-1\right)^2+\left(n-1\right)$$
$$S_{n-1}=2n^2-3n+1$$
$$T_n=S_n-S_{n-1}=2n^2+n-2n^2+3n-1=4n-1$$
$$T_n=4n-1$$
The terms are 3, 7, 11, 15, 19, 23, 27,......
27 is the first term in the series divisible by 9.
27 is the 7th term.
Therefore, the least possible value of n is 7.
The answer is option C.
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Nitin Sahu
2 months ago
By solving the question themselves, they understand themselves. Solving them such a speed as if their train is being missed.
iam jp
3 months ago
Kuch samajh nhi aya