Question

Hello friends here is ur question...
in an ap if s3=24 and s4=30 then find a4 ...
please friends solve it fast it's urgent please friends ...​

Answer 1

$\textbf{Given:}$

$\text{In an A.P, S_3=24 and S_4=30 }$

$\textbf{To find:}$

$a_4$

$\textbf{Solution:}$

$\textbf{Formula used}$

$\text{The n th term of the A.P a, a+d, a+2d, ....... is}$

$\boxed{\bf\,t_n=a+(n-1)d}$

$\text{The sum of n terms of the A.P a, a+d, a+2d,.... is}$

$\boxed{\bf\,S_n=\frac{n}{2}[2a+(n-1)d]}$

$S_3=24$

$\frac{3}{2}[2a+2d]=24$

$\frac{1}{2}[2(a+d)]=8$

$\implies\bf\,a+d=8$---------(1)

$S_4=30$

$\frac{4}{2}[2a+3d]=30$

$2[2a+3d]=30$

$\implies\bf\,2a+3d=15$------(2)

$\text{Using (1) in (2), we get}$

$2a+3(8-a)=15$

$2a+24-3a=15$

$-a=-9$

$\implies\bf\,a=9$

$\text{(1) gives}$

$9+d=8$

$\implies\bf\,d=-1$

$\text{Now,}$

$a_4$

$=\text{4 th term of the A.P}$

$=a+3d$

$=9+3(-1)$

$=9-3$

$=6$

$\implies\boxed{\bf\,a_4=6}$

$\text{}$