Question

The 4th and the last terms of an A.P. are 11 and 89 respectively. If there are 30 terms in the A.P., find the A.P. and its 23rd term.​

Answer 1

`\underline{\underline{\large\bf{Given:-}}}`

`\red{\leadsto}\:`
`\textsf{}`
`\sf Number \: of \:terms \: in \: A.P,n = 30`

`\red{\leadsto}\:`
`\textsf{}`
`\sf Fourth \: term ,a_4 = 11`

`\red{\leadsto}\:`
`\textsf{}`
`\sf last\:term, a_(30) = 89`

`\underline{\underline{\large\bf{To Find:-}}}`

`\orange{\leadsto}\:`
`\textsf{ }`
`\sf The \: A.P.`

`\orange{\leadsto}\:`
`\textsf{ }`
`\sf 23rd\: term, a_(23)`

`\\`

`\underline{\underline{\large\bf{Solution:-}}}\\`

The nth term of A.P is determined by the formula-

`\green{ \underline { \boxed{ \sf{a_n = a+(n-1)d}}}}`

where

• 
  `\sf a = first \:term`
• 
  `\sf a_n = nth \: term`
• 
  `\sf n = number \:of \:terms`
• 
  `\sf d = common \: difference`

Since ,

`\sf a_4 = 11`

`\longrightarrow`
`\sf a+(4-1) d= 11`

`\longrightarrow`
`\sf a+3d= 11\_\_\_(1)`

`\sf a_(30)= 89`

`\longrightarrow`
`\sf a+(30-1)d=89`

`\longrightarrow`
`\sf a+29d= 89\_\_\_(2)`

Subtracting equation (1) from equation(2)

`\begin{gathered}\\\implies\quad \sf a+29d-(a+3d) = 89-11 \\\end{gathered}`

`\begin{gathered}\\\implies\quad \sf a+29d-a-3d = 78 \\\end{gathered}`

`\begin{gathered}\\\implies\quad \sf a-a+29d-3d = 78 \\\end{gathered}`

`\begin{gathered}\\\implies\quad \sf 26d = 78 \\\end{gathered}`

`\begin{gathered}\\\implies\quad \sf d = (78)/(26) \\\end{gathered}`

`\begin{gathered}\\\implies\quad \sf d = 3 \\\end{gathered}`

Putting the value of d in equation (1) -

`\begin{gathered}\\\implies\quad \sf a+3(3) = 11 \\\end{gathered}`

`\begin{gathered}\\\implies\quad \sf a = 11-9 \\\end{gathered}`

`\begin{gathered}\\\implies\quad \sf a = 2 \\\end{gathered}`

• First term of A.P, a = 2

• Second term of A.P.,
  `\sf a_2= 2+(2-1)* 3`

`\quad\quad\quad\sf =2+3`

`\quad\quad\quad\sf =5`

• Third term of A.P.,
  `\sf a_3= 2+(3-1)* 3`

`\quad\quad\quad\sf =2+6`

`\quad\quad\quad\sf =8`

`\longrightarrow`Thus , The A.P is 2,5,8,. . . . . .

Now,

`\begin{gathered}\\\implies\quad \sf a_n = a+(n-1)d \\\end{gathered}`

`\begin{gathered}\\\implies\quad \sf a_(23 )= 2+(23-1)* 3 \\\end{gathered}`

`\begin{gathered}\\\implies\quad \sf a_(23) = 2+22 * 3 \\\end{gathered}`

`\begin{gathered}\\\implies\quad \sf a_(23) = 2+66 \\\end{gathered}`

`\begin{gathered}\\\implies\quad \sf a_(23) = 68 \\\end{gathered}`

`\longrightarrow`Thus , 23rd term is 68.