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Question-74142




Question Number 74142 by MASANJAJ last updated on 19/Nov/19
Answered by Rio Michael last updated on 19/Nov/19
 T_7  = a + 6d = 6 −−−(1)  T_(18)  = a + 17d = 22−−−(2)   eqn(2) − eqn(1) ⇒ 11d = 16   d = ((16)/(11))    a + 6d = 6  ⇒ a + 6(((16)/(11))) = 6      a = 6(1−((16)/(11))) = ((−30)/(11))  S_n  = (1/2)n[ 2a + (n−1)d] = (1/2)n[−((60)/(11)) + (n−1)((16)/(11))]= 252    512 = −((60)/(11))n + ((16)/(11)) n −((16)/(11))  5648 = −44n   n ∉ Z^+
$$\:{T}_{\mathrm{7}} \:=\:{a}\:+\:\mathrm{6}{d}\:=\:\mathrm{6}\:−−−\left(\mathrm{1}\right) \\ $$$${T}_{\mathrm{18}} \:=\:{a}\:+\:\mathrm{17}{d}\:=\:\mathrm{22}−−−\left(\mathrm{2}\right) \\ $$$$\:{eqn}\left(\mathrm{2}\right)\:−\:{eqn}\left(\mathrm{1}\right)\:\Rightarrow\:\mathrm{11}{d}\:=\:\mathrm{16} \\ $$$$\:{d}\:=\:\frac{\mathrm{16}}{\mathrm{11}} \\ $$$$\:\:{a}\:+\:\mathrm{6}{d}\:=\:\mathrm{6} \\ $$$$\Rightarrow\:{a}\:+\:\mathrm{6}\left(\frac{\mathrm{16}}{\mathrm{11}}\right)\:=\:\mathrm{6} \\ $$$$\:\:\:\:{a}\:=\:\mathrm{6}\left(\mathrm{1}−\frac{\mathrm{16}}{\mathrm{11}}\right)\:=\:\frac{−\mathrm{30}}{\mathrm{11}} \\ $$$${S}_{{n}} \:=\:\frac{\mathrm{1}}{\mathrm{2}}{n}\left[\:\mathrm{2}{a}\:+\:\left({n}−\mathrm{1}\right){d}\right]\:=\:\frac{\mathrm{1}}{\mathrm{2}}{n}\left[−\frac{\mathrm{60}}{\mathrm{11}}\:+\:\left({n}−\mathrm{1}\right)\frac{\mathrm{16}}{\mathrm{11}}\right]=\:\mathrm{252} \\ $$$$\:\:\mathrm{512}\:=\:−\frac{\mathrm{60}}{\mathrm{11}}{n}\:+\:\frac{\mathrm{16}}{\mathrm{11}}\:{n}\:−\frac{\mathrm{16}}{\mathrm{11}} \\ $$$$\mathrm{5648}\:=\:−\mathrm{44}{n} \\ $$$$\:{n}\:\notin\:\mathbb{Z}^{+} \\ $$$$\: \\ $$
Commented by MASANJAJ last updated on 19/Nov/19
Thanks sir
Commented by MASANJAJ last updated on 19/Nov/19

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