Question

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

Answer 1

Answer:

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Answer 2

$Let \: \pink{a} \:and \: \blue{b} \:are \\first \:term\: and \: Common \: differnce \:of \:an \:A.P$

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We know that ,

$\boxed { \pink { n^{th} \:term \:of \:A.P = a+(n-1)d }}$

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$i ) 3^{rd} \:term = 12\: (given)$

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

$ii ) last \:term(a_{50}) = 188$

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

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

$\implies a + 49d - (a+2d) = 188 - 12$

$\implies a + 49d - a-2d= 188 - 12$

$\implies 47d = 176$

$\implies d = \frac{176}{47} \: --(3)$

/* Put value of d ,in equation (1) , we get */

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

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

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

$\implies a = \frac{564 -352}{47}$

$\implies a = \frac{212}{47} \:--(4)$

$Now, \red{ 39^{th} \:term } \\= a + 39d \\= \frac{212}{47} + 39 \times \frac{176}{47}\\= \frac{212}{47} + \frac{6864}{47}\\= \frac{212+6864}{47} \\= \frac{7076}{47}$

Therefore.,

$\red{ 39^{th} \:term } \green {= \frac{7076}{47}}$

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