Question

5.
Two A.P.'s are given 9, 7, 5, ... and 24, 21, 18, . . . . If nth term of both the
progressions are equal then find the value of n and nth term.​

Answer 1

hope this answer will be helpful ^_^

Answer 2

AnswEr :

$\bullet\:\textsf{First AP = 9, 7, 5...}\\ \quad \sf a_1 = 9 \:and \:d_1 = (7 - 9) = - 2 \\\\\bullet\:\textsf{Second AP = 24, 21, 18...} \\\quad \sf a_2 = 24 \:and \:d_2 = (24 - 21) = - 3$

$\bigstar\:\boxed{\sf T_n=a + (n - 1)d}$

$\rule{150}{1}$

$\underline{\bigstar\:\textsf{According to the Question Now :}}$

$:\implies\sf T_n\:of\:First\:AP = T_n\:of\:Second\:AP\\\\\\:\implies\sf [a_1 +(n - 1)d_1] = [a_2 +(n - 1)d_2]\\\\\\:\implies\sf [9 +(n - 1) \times ( - 2)] = [24+(n - 1) \times ( - 3)]\\\\\\:\implies\sf [9- 2n + 2] = [24 - 3n + 3]\\\\\\:\implies\sf11 - 2n = 27 - 3n\\\\\\:\implies\sf3n - 2n = 27 - 11\\\\\\:\implies\boxed{ \green{\sf n = 16}}$

⠀

$\therefore\:\underline{\textsf{Value of n in these APs is \textbf{16}}}$

$\rule{200}{2}$

$\underline{\bigstar\:\textsf{Nth term of the AP :}}$

$\longrightarrow\sf T_n = a + (n - 1)d\\\\\\\longrightarrow\sf T_n = 9 + [(16 - 1) (-2)]\\\\\\\longrightarrow\sf T_n = 9 + [15 (-2)]\\\\\\\longrightarrow\sf T_n = 9 - 30\\\\\\\longrightarrow \boxed{ \blue{\sf T_n = -21}}\\\\\qquad\scriptsize{\bf{\dag}\:\texttt{Or, we can alternatively find this as :}}\\\\\longrightarrow\sf T_n = a + (n - 1)d\\\\\\\longrightarrow\sf T_n = 24 + [(16 - 1) (-3)]\\\\\\\longrightarrow\sf T_n = 24 + [15 (-3)]\\\\\\\longrightarrow\sf T_n = 24 - 45\\\\\\\longrightarrow \boxed{\blue{\sf T_n = -21}}$

⠀

$\therefore\:\underline{\textsf{Value of nth term of these APs is \textbf{-21}}}$