Question

The 7th term of an AP is -1 and its 16th term is 17. The nth term of the AP is (a) (3n + 8) (b) (4n - 7) (c) (15 - 2n) (d) (2n - 15)

Answer 1

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

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

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

$Given \: 7^{th} \:term = -1$

$\implies a + 6d = - 1 \: ---(1)$

$16^{th} \:term = 17$

$\implies a + 15d = 17 \: ---(2)$

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

$\implies 9d = 18$

$\implies d = \frac{18}{9}$

$\implies d = 2$

/* Put d = 2 in equation (1), we get */

$a + 6 \times 2 = - 1$

$\implies a + 12 = -1$

$\implies a = - 1 - 12$

$\implies a = - 13$

$Now, \red{ n^{th} \:term }$

$= a+(n-1)d$

$= -13 + ( n - 1 ) \times 2$

$= - 13 + 2n - 2$

$= 2n - 15$

Therefore.,

$\red{ n^{th} \:term }\green { = (2n-15)}$

$Option \: \pink { (d) } \:is \: correct$

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