Question

Find the sum of the first 18 terms of an AP 12b, 8b, 4b ............ *



a) 126b

b) -96b

c) 256b

d) -288b

​

Answer 1

$Given \:A.P: 12b , 8b,4b ,\cdot\cdot\cdot$

$First \:term (a_{1}) = 12b$

$Common \: differnce (d) = a_{2} - a_{1} \\= 8b - 12 b \\= -4b$

$\boxed { \pink { Sum \:of \: n \:terms (S_{n}) = \frac{n}{2} [ 2a + (n-1)d ] }}$

$Here, a = 12b, \:d = -4b \:and \:n = 18$

$Sum\:of \: 18 \:terms = \frac{18}{2} [ 2\times 12b + (18-1)(-4b) ] \\= 9[24b + 17 \times (-4b)]\\= 9(24b - 68b) \\= 9 \times (-44b) \\= - 396b$

Therefore.,

$\red{ Given \:options \:are \:wrong }$

$\red{Sum\:of \: 18 \:terms}\\\green {= - 396b}$

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