Question

the 6th and 11th terms of an arithmetic sequence are 38 and 73. a) Find the first term. b) Write the algebraic expression of the sequence. Please write the correct answer.

Answer 1

$Let \: 'a' \:and \: 'd' \: are \: first \:term\:and$

$Common \: difference \: of\:an \: A.P .$

/* We know that , */

$\boxed{ \pink{ n^{th} \:term (a_{n}) = a + (n-1)d }}$

$Given \: 6^{th} \: term = 38$

$\implies a_{6} = 38$

$\implies a + 5d = 38 \: ---(1)$

$Given \: 11^{th} \: term = 73$

$\implies a_{11} = 73$

$\implies a + 10d = 73 \: ---(2)$

/* Subtract equation (1) from equation (2) , we get */

$\implies 5d = 35$

$\implies d = \frac{35}{5}$

$\implies d = 7$

/* Put d = 7 in equation (1) ,we get */

$a + 5 \times 7 = 38$

$\implies a + 35 = 38$

$\implies a = 38 - 35$

$\implies a = 3$

$\pink {\therefore First \:term (a) = 3 }$

$Here, a = 3, d = 7$

$n^{th} \:term (a_{n}) = 3 + ( n - 1 )\times 7$

$= 3 + 7n - 7$

$= 7n - 4$

$\red { Required \: algebraic \: expression }$

$\green { = 7n - 4 }$

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