Question

If The 9th and 21st term of an A.P are 75 and 83 respectively. Find its 81st term​

Answer 1

Answer:

123

Step-by-step explanation:

let a be the first term and d be the common difference:

given:

a+8d= 75 and a+20d= 83

therefore, a= 75-8d and a= 83-20d

so, 75-8d= 83-20d

or, 20d-8d= 83-75

or, 12d= 8

so, d= 8/12= 2/3

thus, a= 75-8(2/3)= 209/3

hence, 81st term = a+ 80d = 209/3+ 80(2/3) = 369/3

therefore, 81st term = 123

Answer 2

Step-by-step explanation:

$Given \: a_{9} = 75 \: and \\ a_{21} \: = 83 \\ Since, \: \: a_{n1} + ( n_{2} - n_{1})d = a_{n2} \\ \: \: = > a_{9} + (21 - 9)d = a_{21} \\ = > 75 + 12d = 83 \\ = > 12d \: = 83 - 75 \\ 12d \: = 8 \\ = > d = \frac{8}{12} \\ = \frac{2}{3}$

$Then \: a + (n-1)d = a_{n} \\ => a + (9-1) \frac{2}{3} = a_{9} \\ = a + \frac{16}{9} = 75</p><p>\\ a= 75 - \frac{16}{9} \\ a = /frac{209}{3}</p><p>\\ Therefore\: a_{81}= a + (n-1)d</p><p>\\ = \frac{209}{3} + (81-1)\frac{2}{3}</p><p>\\ = \frac{209}{3} + \frac{160}{3}</p><p>\\ = \frac{209+160}{3}</p><p>\\ = \frac{369}{3}</p><p>\\ = 123$

$81^{st} term\: of \: given \: AP \: would\: be \: 123$