Solve the following

Question:

Let Sn and S'4 denote respectively the sum and the sum of the squares of first n natural numbers. If $a_{n}=\frac{S_{n}{ }^{\prime}}{S_{n}}, n \in N .$ Then $\mathrm{a}_{1}, \mathrm{a}_{2}, \mathrm{a}_{3}$, __________$a_{n,}$_____________ forms  an ______ with ______ .

Solution:

Sn : The sum of first n natural numbers .

S: Sum of squares of first natural numbers. 

If $a_{n}=\frac{S_{n}}{S_{n}} ; n \in N$.

$a_{1}=\frac{1}{1}=1$

$a_{2}=\frac{1^{2}+2^{2}}{1+2}=\frac{5}{3}$

$a_{3}=\frac{1^{2}+2^{2}+3^{2}}{1+2+3}=\frac{14}{6}=\frac{7}{3} \ldots \ldots a_{n}=\frac{1^{2}+2^{2}+\ldots .+n^{2}}{1+2+\ldots .}$

$=\frac{\frac{n(n+1)(2 n+1)}{6}}{\frac{n(n+1)}{2}}$

$=\frac{(2 n+1)}{6} \times 2$

$a_{n}=\frac{S_{n}{ }^{\prime}}{S_{n}}=\frac{2 n+1}{3}$

Hence, common difference is

$a_{2}-a_{1}=\frac{5}{3}-1=\frac{2}{3}$

$a_{3}-a_{2}=\frac{7}{3}-\frac{5}{3}=\frac{2}{3}$

Hence, common difference of $A$.? is $\frac{2}{3}$.

 

 

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