$\sqrt{2}, \sqrt{8}, \sqrt{18}$ …… forms a progression.
(a) Identify the type of progression.
(b) The sum of its first 10 terms is $p\sqrt{2}$. Find the value of $p$.
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(a) Given,
$\sqrt{2}, \sqrt{8}, \sqrt{18}$ ......
Simplifying the terms,
$\sqrt{2}, 2\sqrt{2}, 3\sqrt{2}$ ......
Second term − First term = $2\sqrt{2} - \sqrt{2} = \sqrt{2}$
Third term − Second term = $3\sqrt{2} - 2\sqrt{2} = \sqrt{2}$
Since the difference between consecutive terms is constant, the progression is an Arithmetic Progression (A.P.) with first term $a = \sqrt{2}$ and common difference $d = \sqrt{2}$.
Hence, the progression is an Arithmetic Progression.
(b) Given,
$a = \sqrt{2}, d = \sqrt{2}, n = 10$
By formula,
Sum of first $n$ terms = $\frac{n}{2}[2a + (n - 1)d]$
Substituting values we get :
$$\begin{aligned}= \frac{10}{2}[2\sqrt{2} + (10 - 1)\sqrt{2}] \\[1em] = 5[2\sqrt{2} + 9\sqrt{2}] \\[1em] = 5 \times 11\sqrt{2} \\[1em] = 55\sqrt{2}.\end{aligned}$$
Given, the sum of first 10 terms = $p\sqrt{2}$.
$$\begin{aligned}\Rightarrow p\sqrt{2} = 55\sqrt{2} \\[1em] \Rightarrow p = 55.\end{aligned}$$
Hence, the value of $\bm{p}$ = 55.