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For each positive integer $n$, the mean of first $n$ terms is $n$. What is the 17th term of the sequence? Now, how to find the sum since any particular series is not specified?

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    Any thoughts? What's $a_1$? What's $a_2$? – 2017-01-16
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    @lulu nope I don't know! What will it be? – 2017-01-16
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    What's the mean of the first two terms? What's... er... the mean of the first **one** term(s)? – 2017-01-16

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Hint

$$\frac{a_1+a_2+...+a_n}{n}=n \rightarrow S=a_1+a_2+...+a_n=n^2$$

So,

$$a_{17}=(a_1+a_2+...+a_{17})-(a_1+a_2+...+a_{16})$$

Can you finish?

For a general approach you can write:

$$a_n=(a_1+a_2+...+a_{n})-(a_1+a_2+...+a_{n-1})=n^2-(n-1)^2=2n-1$$

And the sum is of course given by:

$$S=a_1+a_2+...+a_{n}=n^2$$

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    @Lokesh Sangewar: is it clear? – 2017-01-18
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We have $$\frac{\sum_{k=1}^n a_k}{n} = n$$ so: $$\sum_{k=1}^n a_k = n^2$$ $$\sum_{k=1}^{n-1} a_k = (n-1)^2$$ $$a_n = \sum_{k=1}^n a_k - \sum_{k=1}^{n-1} a_k = n^2-(n-1)^2 = (n+n-1)(n-n+1)=2n-1$$

We have then $$a_{17}=2\cdot 17-1 = 33$$

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    How did u get that last expression An=? – 2017-01-16
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    @LokeshSangewar $\sum_{k=1}^na_k = \sum_{k=1}^{n-1}a_k+a_n$ – 2017-01-16
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Let the $r$-th term, $a_r=f(r)$.

Given that the average of the first $n$ terms is $n$, we have $$\begin{align} \frac 1n\sum_{r=0}^n f(r)&=n\\ \sum_{r=0}^n f(r)&=n^2\\ f(n)&=n^2-(n-1)^2=2n-1\\ f(17)&=\color{red}{33}\end{align}$$