Question

Tell me how to derive this Sn=u+a(n-1)​

Answer 1

Correct question:-

How to derive $\bold{S_n = u + \frac{a}{2}[2n - 1]}$

$\huge{\bold{\underline{\underline{....Answer....}}}}$

$\huge{\bold{\underline{Given:-}}}$

Let's assume some points.

u = initial velocity.

a = acceleration.

$\huge{\bold{\underline{Explanation:-}}}$

Distance covered in (n) seconds.

From second kinematic equation

$\large{\bold{S = ut + \frac{1}{2}at^2}}$

Substituting the values.

$\large{ \implies S_1 = u(n) + \frac{1}{2}a(n)^2}$

$\large{\boxed{ \therefore S_1 = un + \frac{1}{2}an^2}}$

Distance covered in (n - 1) seconds.

From Second kinematic equation.

$\large{\bold{S_2 = ut + \frac{1}{2} at^2}}$

Substituting the values.

$\large{ \implies S_2 = u(n - 1) + \frac{1}{2}a (n - 1)^2}$

$\large{\boxed{ \therefore S_2 = u(n - 1) + \frac{1}{2}a (n^2 + 1 - 2n)}}$

To get Sn we need to

$\large{\bold{S_n = S_1 - S_2}}$

Substituting the values.

$\large{ \implies S_n = \left[un + \frac{1}{2}an^2 \right] - \left[ un - u + \frac{1}{2}a (n^2 + 1 - 2n) \right]}$

Simplifying

$\large{ \implies S_n = \cancel{un} + \cancel{\frac{1}{2} an^2} - \cancel{un} + u - \cancel{ \frac{1}{2} an^2} - \frac{1}{2}a[1 - 2n]}$

Now,

$\large{ \implies S_n = u + \frac{1}{2}a[2n - 1]}$

$\huge{\boxed{\boxed{S_n = u + \frac{a}{2}[2n - 1]}}}$

Hence derived.