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Question Number 127360 by bemath last updated on 29/Dec/20

	Answered by liberty last updated on 29/Dec/20
	$let : a−7b,a−6b,a−5b,..., a+5b, a+6b, a+7b is AP given condition → { ((3a−18b=−60; a−6b=−20 (the first 3 terms))),((3a+18b=84; a+6b = 28 (the last 3 terms))) :} we get { ((a=4)),((b=4)) :}. we want to compute the sum of the middle 3 terms ⇒T_7 +T_8 +T_9 = (a−b)+a+(a+b) = 3a = 3×4=12$
	$l e t : a - 7 b, a - 6 b, a - 5 b, . . ., a + 5 b, a + 6 b, a + 7 b i s A P$ $g i v e n c o n d i t i o n \to {\begin{cases} 3 a - 18 b = - 60; a - 6 b = - 20 (t h e f i r s t 3 t e r m s) \\ 3 a + 18 b = 84; a + 6 b = 28 (t h e l a s t 3 t e r m s) \end{cases}$ $w e g e t {\begin{cases} a = 4 \\ b = 4 \end{cases} . w e w a n t t o c o m p u t e t h e$ $s u m o f t h e m i d d l e 3 t e r m s \Rightarrow T_{7} + T_{8} + T_{9}$ $= (a - b) + a + (a + b) = 3 a = 3 \times 4 = 12$
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