Question

Established the formula Sn =n/2[2a+(n-1)d] for the sum of the first n terms of an AP.​

Answer 1

$\large\underline{\sf{Solution-}}$

Let us consider an AP series whose

• First term is a

• Common difference is d

• Number of terms is n.

$\sf \: Let \: a_1, \: a_2, \: a_3, - - - ,a_{n - 1}, \: a_n \: be \: n \: terms \: of \: AP$

and

$\sf \: Let \: S_n \: represents \: sum \: of \: n \: terms \: of \: AP$

So,

$\rm :\longmapsto\:S_n = a_1 + a_2 + - - - + a_{n - 1} + a_n - - - (1)$

can also be rewritten as

$\rm :\longmapsto\:S_n = a_n + a_{n - 1} + - - - + a_2 + a_1 - - - (2)$

On adding equation (1) and equation (2), we get

$\rm :\longmapsto\:2S_n = (a_1 + a_n)+(a_2 + a_{n - 1}) + - - + (a_{n - 1} + a_2) + (a_n+ a_1)$

Wᴇ ᴋɴᴏᴡ ᴛʜᴀᴛ,

↝ nᵗʰ term of an arithmetic sequence is,

$\begin{gathered}\red\bigstar\:\:{\underline{\orange{\boxed{\bf{\green{a_n\:=\:a\:+\:(n\:-\:1)\:d}}}}}} \\ \end{gathered}$

Wʜᴇʀᴇ,

aₙ is the nᵗʰ term.

a is the first term of the sequence.

n is the no. of terms.

d is the common difference.

Tʜᴜs,

$\rm :\longmapsto\:a_1 = a$

$\rm :\longmapsto\:a_2 = a + d$

$\rm :\longmapsto\:a_3 = a + 2d$

.

.

.

.

.

$\rm :\longmapsto\:a_{n - 1} = a + (n - 2)d$

$\rm :\longmapsto\:a_{n} = a + (n - 1)d$

So,

$\rm :\longmapsto\:a_1 + a_{n} = a + a + (n - 1)d = 2a + (n - 1)d$

$\rm :\longmapsto\:a_2 + a_{n - 1} = a + d + a + (n - 2)d = 2a + (n - 1)d$

So, on substituting all these values, we get

$\rm :\longmapsto\:2S_n = \bigg(2a + (n - 1)d\bigg) + \bigg(2a + (n - 1)d\bigg) + - - - + \bigg(2a + (n - 1)d\bigg)$

$\rm :\longmapsto\:2S_n =n \bigg(2a + (n - 1)d\bigg)$

$\bf\implies \: \: S_n= \: \dfrac{n}{2}\bigg(2a + (n - 1)d\bigg)$

Hence, Proved

Answer 2

Step-by-step explanation:

S=n/2(2a+(n-1)d find a here