Q.1 Find the sum of 𝑛 terms of the series

$(4-\frac{1}{n}) + (4 - \frac{2}{n}) + (4 - \frac{3}{n})+.....$

 

(OR)

Q.2 If the roots of the equation  (𝑎² + 𝑏² )𝑥²   − 2(𝑎𝑐 + 𝑏𝑑)𝑥 + (𝑐²   + 𝑑² ) are equal, prove that $\frac{a}{b}=\frac{c}{d}$            

We need to find the sum of 𝑛 terms of the series

$(4-\frac{1}{n}) + (4 - \frac{2}{n}) + (4 - \frac{3}{n})+.....$

It can be observed that the series forms an AP with common difference

$(4 - \frac{2}{n}) - (4 - \frac{1}{n}) = -\frac{1}{n}$

It is known that the sum of an AP is given by

𝑆_n  = 𝑛/2[2𝑎 + (𝑛 − 1)𝑑], where 𝑆_n is the sum of terms, 𝑛 is the number of terms, 𝑎 is the first term and 𝑑 is the common difference of AP. On substituting the values, we get
$$\begin{gathered} S_{n = }\frac{n}{2}\left[2(4 -\frac{1}{n})+(n-1)(-\frac{1}{n})\right] \\ {S}_n = \frac{n}{2}\left[8 - \frac{2}{n}- 1 + \frac{1}{n}\right] \\ {S}_n = \frac{n}{2}\left[7-\frac{1}{n}\right] \end{gathered}$$

Hence, the sum of the series is $\frac{n}{2}\left[7 -\frac{1}{n}\right]$

(OR)

If the roots of the equation $\frac{a}{b}=\frac{c}{d}$ provided the roots of the equation
(𝑎² + 𝑏² )𝑥²   − 2(𝑎𝑐 + 𝑏𝑑)𝑥 + (𝑐²   + 𝑑² ) = 0 are equal.

It can be observed that the above equation is a quadratic equation. On comparing it with the general form of quadratic equation 𝑝𝑥² + 𝑞𝑥 + 𝑟 = 0, where 𝑝, 𝑞 and 𝑟 are constants, we get
𝑝 = 𝑎 + 𝑏
𝑞 =− 2(𝑎𝑐 + 𝑏𝑑)
𝑟 = (𝑐² + 𝑑² )

It is given that the roots are equal therefore the discriminant (𝐷) must be zero.
𝐷 = $\sqrt{q-4pr}$ = 0 or
𝐷 = 𝑞   − 4𝑝𝑟 = 0
On substituting the values, we get
⇒ 𝑞² − 4𝑝𝑟 = 0
⇒ 𝑞² = 4𝑝𝑟
⇒ [− 2(𝑎𝑐 + 𝑏𝑑)]² = 4(𝑎²   + 𝑏² )(𝑐²   + 𝑑² )
⇒ 4(𝑎𝑐 + 𝑏𝑑) = 4(𝑎²   + 𝑏² )(𝑐²   + 𝑑² )
On simplifying, we get
⇒ 2𝑎𝑏𝑐𝑑 = 𝑎² 𝑑² + 𝑏² 𝑐²
⇒ 𝑎 𝑑 + 𝑏 𝑐   − 2𝑎𝑏𝑐𝑑 = 0
⇒ (𝑎𝑑 − 𝑏𝑐) = 0
⇒ 𝑎𝑑 − 𝑏𝑐 = 0
⇒ 𝑎/b   = 𝑐/d 
Hence, proved