Answer :
[tex]\underline{\underline{\large\bf{Given:-}}}[/tex]
[tex]\red{\leadsto}\:[/tex][tex]\textsf{} [/tex][tex]\sf Number \: of \:terms \: in \: A.P,n = 30[/tex]
[tex]\red{\leadsto}\:[/tex][tex]\textsf{} [/tex][tex]\sf Fourth \: term ,a_4 = 11 [/tex]
[tex]\red{\leadsto}\:[/tex][tex]\textsf{} [/tex][tex]\sf last\:term, a_{30} = 89 [/tex]
[tex]\underline{\underline{\large\bf{To Find:-}}}[/tex]
[tex]\orange{\leadsto}\:[/tex][tex]\textsf{ } [/tex][tex]\sf The \: A.P. [/tex]
[tex]\orange{\leadsto}\:[/tex][tex]\textsf{ } [/tex][tex]\sf 23rd\: term, a_{23} [/tex]
[tex]\\[/tex]
[tex]\underline{\underline{\large\bf{Solution:-}}}\\[/tex]
The nth term of A.P is determined by the formula-
[tex]\green{ \underline { \boxed{ \sf{a_n = a+(n-1)d}}}}[/tex]
where
- [tex]\sf a = first \:term[/tex]
- [tex]\sf a_n = nth \: term[/tex]
- [tex]\sf n = number \:of \:terms [/tex]
- [tex]\sf d = common \: difference [/tex]
Since ,
[tex]\sf a_4 = 11 [/tex]
[tex]\longrightarrow[/tex] [tex]\sf a+(4-1) d= 11 [/tex]
[tex]\longrightarrow[/tex] [tex]\sf a+3d= 11\_\_\_(1) [/tex]
[tex]\sf a_{30}= 89 [/tex]
[tex]\longrightarrow[/tex] [tex]\sf a+(30-1)d=89[/tex]
[tex]\longrightarrow[/tex][tex]\sf a+29d= 89\_\_\_(2)[/tex]
Subtracting equation (1) from equation(2)
[tex]\begin{gathered}\\\implies\quad \sf a+29d-(a+3d) = 89-11 \\\end{gathered} [/tex]
[tex]\begin{gathered}\\\implies\quad \sf a+29d-a-3d = 78 \\\end{gathered} [/tex]
[tex]\begin{gathered}\\\implies\quad \sf a-a+29d-3d = 78 \\\end{gathered} [/tex]
[tex]\begin{gathered}\\\implies\quad \sf 26d = 78 \\\end{gathered} [/tex]
[tex]\begin{gathered}\\\implies\quad \sf d = \frac{78}{26} \\\end{gathered} [/tex]
[tex]\begin{gathered}\\\implies\quad \sf d = 3 \\\end{gathered} [/tex]
Putting the value of d in equation (1) -
[tex]\begin{gathered}\\\implies\quad \sf a+3(3) = 11 \\\end{gathered} [/tex]
[tex]\begin{gathered}\\\implies\quad \sf a = 11-9 \\\end{gathered} [/tex]
[tex]\begin{gathered}\\\implies\quad \sf a = 2 \\\end{gathered} [/tex]
- First term of A.P, a = 2
- Second term of A.P.,[tex]\sf a_2= 2+(2-1)\times 3[/tex]
[tex]\quad\quad\quad\sf =2+3[/tex]
[tex]\quad\quad\quad\sf =5 [/tex]
- Third term of A.P.,[tex]\sf a_3= 2+(3-1)\times 3[/tex]
[tex]\quad\quad\quad\sf =2+6[/tex]
[tex]\quad\quad\quad\sf =8 [/tex]
[tex]\longrightarrow[/tex]Thus , The A.P is 2,5,8,. . . . . .
Now,
[tex]\begin{gathered}\\\implies\quad \sf a_n = a+(n-1)d \\\end{gathered} [/tex]
[tex]\begin{gathered}\\\implies\quad \sf a_{23 }= 2+(23-1)\times 3 \\\end{gathered} [/tex]
[tex]\begin{gathered}\\\implies\quad \sf a_{23} = 2+22 \times 3 \\\end{gathered} [/tex]
[tex]\begin{gathered}\\\implies\quad \sf a_{23} = 2+66 \\\end{gathered} [/tex]
[tex]\begin{gathered}\\\implies\quad \sf a_{23} = 68 \\\end{gathered} [/tex]
[tex]\longrightarrow[/tex]Thus , 23rd term is 68.