Question

Hey mate.......answer it ....☺☺☺☺✌✌✌​

Answer 1

Given :

☢  First and Last term of the A.P are a and l

☢  Sum of the A.P is S

Let the Number of terms in the A.P be : n

Let the common difference be : d

We know that - nth term in an A.P is given by :

$\large\boxed{\mathsf{\bigstar\;\;T_n = a + (n - 1)d}}}$

As nth term is equal to last term :

$\implies \textsf{l = a + (n - 1)d}$

$\implies \textsf{l - a = (n - 1)d}$

$\implies \mathsf{d = \dfrac{l - a}{n - 1}}$

$\mathsf{Given : common\;difference\;(d) = \dfrac{l^2 - a^2}{k - (l + a)}}$

$\mathsf{\implies \dfrac{l^2 - a^2}{k - (l + a)} = \dfrac{l - a}{n - 1}}$

$\mathsf{\implies \dfrac{(l - a)(l + a)}{k - (l + a)} = \dfrac{l - a}{n - 1}}\\\\\\\mathtt{As : (l^2 - a^2) = (l + a)(l - a)}}$

$\mathsf{\implies \dfrac{l + a}{k - (l + a)} = \dfrac{1}{n - 1}}$

$\mathsf{\implies (l + a)(n - 1) = k - (l + a)}$

$\mathsf{\implies k = (l + a)(n - 1) + (l + a)}$

$\mathsf{\implies k = (l + a)[n - 1 + 1]}$

$\mathsf{\implies k = (l + a)(n)}$

We know that - Sum of 'n' terms in an A.P is given by :

$\mathsf{\bigstar\;\;S = \dfrac{n}{2}[2a + (n - 1)d]}$

$\mathsf{\implies S = \dfrac{n}{2}[a + a + (n - 1)d]}$

$\mathsf{\implies S = \dfrac{n}{2}[a + l]}$

$\mathsf{\implies (n)(a + l) = 2S}$

But, We found that (n)(a + l) = k

$\mathsf{\implies k = 2S}$

$\underline{\textbf{Answer}} : \textsf{k is equal to = \large\boxed{\mathsf{2S}}}$