Question

in an ap the sum of 25 terms is 400 then the 13 the terms is​

Answer 1

Question:

In an AP the sum of 25 terms is 400 then the 13 the terms is.

Given:

Sum of 25 terms, $\mathsf{{S}_{25}=400}$

To Find:

$\mathsf{13}^{th}$term of AP.

Formula to be used:

$\mathsf{{S}^{n}=\large\frac{n}{2}[2a+(n-1)d]}$

Solution:

$\dashrightarrow\mathsf{{S}^{n}=\large\frac{n}{2}[2a+(n-1)d]}$

$\dashrightarrow\mathsf{400=\large\frac{25}{2}[2a+(25-1)d]}$

$\dashrightarrow\mathsf{400=\large\frac{25}{2}[2a+(24)d]}$

$\dashrightarrow\mathsf{400\times 2=25[2a+24d]}$

$\dashrightarrow\mathsf{800=25[2a+24d]}$

$\dashrightarrow\mathsf{\large\frac{800}{25}=2a+24d}$

$\dashrightarrow\mathsf{32=2a+24d}$

$\dashrightarrow\mathsf{32=2(a+12d)}$

$\dashrightarrow\mathsf{\large\frac{32}{2}=a+12d}$

$\dashrightarrow\mathsf{16=a+12d}$

We know, $\mathsf{a+12d={a}_{13}}$. So,

$\color{red}\dashrightarrow\mathsf{{a}_{13}=16}$

$\therefore\: \mathsf{13}_{th}$ term of the Arithmetic progression is 16.

$\underline\color{skyblue}\mathsf{More\: to\: know:}$

-The sum to n terms of an AP with first term 'a' and common difference 'd' is given by:

$\mathsf{{S}^{n}=\large\frac{n}{2}[2a+(n-1)d]}$

Also, $\mathsf{{S}_{n}=\large\frac{n}{2}[a+l]}$, where,

l= last term =a+(n-1)d