CAT 2022 Question Paper (Slot 1) Question 47

Question 47

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

Solution

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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Comments

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

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