Question

The 17th term of an ap is 5more than twice its 8th term if 11 th term of ap is 43 find nth term

Answer 1

Hi Friend,

a₁₇ = 5 + 2a₈                  and a₁₁ = 43 
APPLY THE NTH TERM FORMULA =  a + (n-1) d    where  a = first term  d = common difference of A.P
                                                  
THERFORE, 

           a + (17  - 1) d  =  5 + 2{ a + (8 - 1 ) d}  

           a + 16 d =  5 + 2a + 14d        2d - a = 5        .......(1)

      a + (11 - 1)d = 43             
a + 10d = 43        .........(2)


SOLVE 1 AND 2     12d = 48              
d = 4 


 a=3



Hope it helps you!!!

Answer 2

Given: 

• The 17th Term of an AP is five more than twice the 8th term of the AP. & The 11th term of the AP is 43.

To find: 

• The first term (a) & the Common difference ( d )?

⠀⠀

Solution:

◕ By using formula = ⟩⟩ aₙ = a + (n – 1) d ⟨⟨

» 8th term of AP —

$\begin{gathered}\longrightarrow\bf a_8 = a + 7d\qquad\qquad\quad\sf\Bigg\lgroup eq^{n}\;(1)\Bigg\rgroup\\\\\end{gathered}$

» 17th term of AP —

$\begin{gathered}\longrightarrow\bf a_{17} = a + 16d\qquad\qquad\quad\sf\Bigg\lgroup eq^{n}\;(2)\Bigg\rgroup\\\\\end{gathered}$

$\begin{gathered}\underline{\maltese \:\boldsymbol{\sf{According\;to\;the\; Question\; :}}}\\\\\end{gathered}$

• It is Given that, 17th term of the AP is five more than twice the 8th term of the AP.

⠀

$\begin{gathered}:\implies\sf a_{17} = 2(a_8) +5\\\\\\\end{gathered}$

$\begin{gathered}:\implies\sf a + 16d = 2(a + 7d) + 5\qquad\qquad\quad\sf\Bigg\lgroup Using\;eq^{n}\;(1)\:\&\;eq^{n}\;(2)\Bigg\rgroup\\\\\\\end{gathered}$

$\begin{gathered}:\implies\sf a + 16d = 2a + 14d + 5\\\\\\\end{gathered}$

$\begin{gathered}:\implies\sf 2a - a = 16d - 14d - 5\\\\\\\end{gathered}$

$\begin{gathered}:\implies\boxed{\frak{a = 2d -5}}\blue\bigstar \qquad\qquad\quad\quad\sf\Bigg\lgroup eq^{n}\;(3)\Bigg\rgroup\\\\\\\end{gathered}$

Second Condition:

• As it is given that, 11th term of the AP is 43.⠀

$\begin{gathered}:\implies\sf a_{11} = a + (11 - 1)d\\\\\\\end{gathered}$

$\begin{gathered}:\implies\sf 43 = a + 10d\\\\\\\end{gathered}$

$\begin{gathered}:\implies\sf 43 = 2d - 5 10d\qquad\qquad\quad\quad\sf\Bigg\lgroup From\;eq^{n}\;(3)\Bigg\rgroup\\\\\\\end{gathered}$$\begin{gathered}:\implies\sf 12d = 43 + 5\\\\\\\end{gathered} \\ \begin{gathered}:\implies\sf 12d = 48\\\\\\\end{gathered} \\ \begin{gathered}:\implies\sf d = \cancel\dfrac{48}{12}\\\\\\\end{gathered}$

$\begin{gathered}:\implies{\underline{\boxed{\frak{ d = 4}}}}\pink\bigstar\\\\\end{gathered}$

• Hence, the term is 4

Thank you!!

@itzshivani