Question

Let S_n and S'₄ denote respectively the sum and the sum of the squares of first n natural numbers. If $a_n=\frac{S_n\text{'}}{S_n},n\in N.$ Then a₁, a₂, a₃,_____, a_n, _____ forms an ______ with ______ .

Answer 1

S_n : The sum of first n natural numbers .
S_n : Sum of squares of first n natural numbers.
If a_n = $\frac{S_n\text{'}}{S_n}$ ; n∊N.
$$\begin{gathered} a_1=\frac{1}{1}=1 \\ {a}_2=\frac{1^2+2^2}{1+2}=\frac{5}{3} \end{gathered}$$

$\begin{array}{rcl}{a}_3& =& \frac{1^2+2^2+3^2}{1+2+3}=\frac{14}{6}=\frac{7}{3}.......a_n=\frac{1^2+2^2+....+n^2}{1+2+....n}\\ & =& \frac{{\displaystyle \frac{n\left(n+1\right)\left(2n+1\right)}{6}}}{{\displaystyle \frac{n\left(n+1\right)}{2}}}\\ & =& \frac{\left(2n+1\right)}{6}\times 2\\ a_n& =& \frac{{S_n}^{\text{'}}}{S_n}=\frac{2n+1}{3}\end{array}$
Hence, common difference is
a₂ − a₁ = $\frac{5}{3}-1=\frac{2}{3}$
a₃ − a₂ = $\frac{7}{3}-\frac{5}{3}=\frac{2}{3}$
Hence, common difference of A.? is $\frac{2}{3}$.