Question

If the 9th and 21 th terms of an A.P.are 75 and183 respectively , find its 81 st term​

Answer 1

let 9d and 21d is a+8d=75 a+20d=183 -12d=-108 d9

Answer 2

$\huge{\underline{\underline{\bf{Solution}}}}$

$\rule{200}{2}$

$\tt given\begin{cases} \sf{9^{th} \: term \: of \: the \: A.P \: is 75} \\ \sf{21^{st} \: term \: of \: the \: A.P \: is \: 183} \end{cases}$

$\rule{200}{2}$

$\Large{\underline{\underline{\bf{To \: Find :}}}}$

We have to find $\sf{81^{st}}$ term of the A.P.

$\rule{200}{2}$

$\Large{\underline{\underline{\bf{Explanation :}}}}$

A.T.Q

a + 8d = 75 .......... (1)

a + 20d = 183 ..........(2)

____________________________

→From equation (1)

a = 75 - 8d

→ Put value of a in equation (2)

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75 - 8d + 20d = 183

→75 + 12d = 183

→12d = 108

→d = 108/12

→d = 9

____________________________

Put Value of d in equation (1)

a + 8(9) = 75

a = 75 - 72

a = 72

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Now,

$\Large{\star{\boxed{\sf{A_n = a + (n - 1)d}}}}$

$\sf{A_{81} = 3 + (81 - 1)3} \\ \\ \sf{A_{81} = 3 + 240} \\ \\ \sf{A_{81} = 243}$

$\sf{\therefore \: 81^{st} \: term \: of \: the \: A.P \: is \: 243}$