Question

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

Answer 1

Answer:

A17=5+2A8

a+16d=5+2a+14d

-a+2d=-5

a-2d=5.........eq.1

a11=43

a+10d=43(given)..........eq.2

solving eq. 1 and 2

a-2d=5

a+10d=43

----------------

8d=38

d=19/4

putting the value of (d) in eq. 1

a-2×19/4=5

a=-9/2

an=a+(n-1)d

an= -9/2+(n-1)19/4

= -27/4+19n/4

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