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^+](https://www.tinkutara.com/question/Q74174.png)
$$\:{T}_{\mathrm{7}} \:=\:{a}\:+\:\mathrm{6}{d}\:=\:\mathrm{6}\:−−−\left(\mathrm{1}\right) \\ $$$${T}_{\mathrm{18}} \:=\:{a}\:+\:\mathrm{17}{d}\:=\:\mathrm{22}−−−\left(\mathrm{2}\right) \\ $$$$\:{eqn}\left(\mathrm{2}\right)\:−\:{eqn}\left(\mathrm{1}\right)\:\Rightarrow\:\mathrm{11}{d}\:=\:\mathrm{16} \\ $$$$\:{d}\:=\:\frac{\mathrm{16}}{\mathrm{11}} \\ $$$$\:\:{a}\:+\:\mathrm{6}{d}\:=\:\mathrm{6} \\ $$$$\Rightarrow\:{a}\:+\:\mathrm{6}\left(\frac{\mathrm{16}}{\mathrm{11}}\right)\:=\:\mathrm{6} \\ $$$$\:\:\:\:{a}\:=\:\mathrm{6}\left(\mathrm{1}−\frac{\mathrm{16}}{\mathrm{11}}\right)\:=\:\frac{−\mathrm{30}}{\mathrm{11}} \\ $$$${S}_{{n}} \:=\:\frac{\mathrm{1}}{\mathrm{2}}{n}\left[\:\mathrm{2}{a}\:+\:\left({n}−\mathrm{1}\right){d}\right]\:=\:\frac{\mathrm{1}}{\mathrm{2}}{n}\left[−\frac{\mathrm{60}}{\mathrm{11}}\:+\:\left({n}−\mathrm{1}\right)\frac{\mathrm{16}}{\mathrm{11}}\right]=\:\mathrm{252} \\ $$$$\:\:\mathrm{512}\:=\:−\frac{\mathrm{60}}{\mathrm{11}}{n}\:+\:\frac{\mathrm{16}}{\mathrm{11}}\:{n}\:−\frac{\mathrm{16}}{\mathrm{11}} \\ $$$$\mathrm{5648}\:=\:−\mathrm{44}{n} \\ $$$$\:{n}\:\notin\:\mathbb{Z}^{+} \\ $$$$\: \\ $$
Commented by MASANJAJ last updated on 19/Nov/19
Thanks sir
Commented by MASANJAJ last updated on 19/Nov/19
