Question

The resistances R,(R+1),(R+2)-------(R+n) are connected in series find equivalent resistance

Answer 1

Answer:

R= R + R +1 + R+ 2 + R + n

R = 4R + 3 + n

R/4R = 3 + n

1/4 = 3 + n

1/4 -3= n

1-12/4 = n

-11/4 = n

I Think This Is The Answer If Not Then Correct Me Also

Answer 2

Total number of Resistors = n

Their Resistances are as follows :-

R, (R+1),(R+2)+.....+(R+n)

$\rule{200}{2}$

Now,

In series Connections we know that

$R_{eq} =R_1+R_2+R_3+....+R_n$

$\rule{200}{2}$

Here also,

Applying the same formula for series connections :-

Thus,

The equivalent Resistance will be equal to :-

$=R+R+1+R+2+R+3+....R+n\\ \\ \\=(R+R.... n\:terms) +(1+2+3+...+n)\\ \\ \\=nR+(\dfrac{n(n+1)}{2})\\ \\ \\=n(\dfrac{(2R+n+1)}{2})$

$\rule{200}{2}$

Now,

For explanation :-

Consider an AP

1,2,3,4...... n terms

a=1 and d=1

Sum of the AP is :-

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

$\rule{200}{2}$

Hence,

Equivalent Resistance is equal to

$=n(\dfrac{(2R+n+1) }{2}$

$\rule{200}{2}$