Question

An
Ap
consists of 50 terms of which 3rd term is 12 and last term is 188 find the 39th term
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Answer 1

$Let \: \pink { a \: and \: d }\: are \: first \:term\\and \: Common \: difference \: of \: an \: A.P.$

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

/* According to the problem given */

$3^{rd} \:term = 12$

$\implies a + 2d = 12 \: --( 1)$

$Last \: term = 188$

$\implies 50^{th} \:term = 188$

$\implies a + 49 d = 188 \: --(2)$

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

$\implies 47d = 176$

$\implies d = \frac{176}{47}$

$Put \: d = \frac{176}{47}\: in \: equation (1) ,\\ we \:get$

$\implies a +2 \times \frac{176}{47} = 12$

$\implies a = 12 - \frac{352}{47}$

$\implies a = \frac{574 - 352}{47}$

$\implies a = \frac{212}{47}$

$Now, \red{39^{th} \:term \: of \: A.P }\\= a + 38d \\= \frac{212}{47} + 38 \times \frac{176}{47}$

$= \frac{ 212 + 6688}{47} \\\green {= \frac{6900}{47}}$

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