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Question Number 22090 by tawa tawa last updated on 10/Oct/17

	Answered by Tikufly last updated on 11/Oct/17

	$(i) 3, 15, 75, 375, . . . . . . .$ $(ii) 3, 15, 27, 39, . . . . . . . . .$
	Commented by tawa tawa last updated on 11/Oct/17

	$God bless you sir$
	Answered by $@ty@m last updated on 11/Oct/17
	$ATQ t_2 ^2 =t_1 .t_7 (a+d)^2 =a(a+6d) a^2 +d^2 +2ad=a^2 +6ad d^2 −4ad=0 d(d−4a)=0 d=4a −−(1) r=(t_2 /t_1 )=((a+d)/a)=((5a)/a)=5 Let product of n terms of G.P.=3375 ⇒A.Ar.Ar^2 ....Ar^(n−1) =3375 ⇒A^n .r^((n(n−1))/2) =3375 ⇒A^n .5^((n(n−1))/2) =3^3 5^3 ⇒n=3, A=3 ∴ the Geometric series is 3+15+75 Ans.(i) Now, t_2 =a+d ⇒Ar=a+4a ⇒15=5a ⇒a=3 &d=12 ∴ using formula S_n =(n/2){2a+(n−1)d} S_n =(n/2){6+(n−1).12} S_n =(n/2)(12n−6) Ans.(ii)$
	$A T Q$ $t_{2}^{2} = t_{1} . t_{7}$ ${(a + d)}^{2} = a (a + 6 d)$ $a^{2} + d^{2} + 2 a d = a^{2} + 6 a d$ $d^{2} - 4 a d = 0$ $d (d - 4 a) = 0$ $d = 4 a - - (1)$ $r = \frac{t_{2}}{t_{1}} = \frac{a + d}{a} = \frac{5 a}{a} = 5$ $L e t p r o d u c t o f n t e r m s o f G . P . = 3375$ $\Rightarrow A . A r . {A r}^{2} . . . . {A r}^{n - 1} = 3375$ $\Rightarrow A^{n} . r^{\frac{n (n - 1)}{2}} = 3375$ $\Rightarrow A^{n} . 5^{\frac{n (n - 1)}{2}} = 3^{3} 5^{3}$ $\Rightarrow n = 3, A = 3$ $∴ t h e G e o m e t r i c s e r i e s i s$ $3 + 15 + 75 A n s . (i)$ $N o w, t_{2} = a + d$ $\Rightarrow A r = a + 4 a$ $\Rightarrow 15 = 5 a$ $\Rightarrow a = 3 & d = 12$ $∴ u s i n g f o r m u l a S_{n} = \frac{n}{2} {2 a + (n - 1) d}$ $S_{n} = \frac{n}{2} {6 + (n - 1) . 12}$ $S_{n} = \frac{n}{2} (12 n - 6) A n s . (i i)$
	Commented by tawa tawa last updated on 11/Oct/17

	$God bless you sir .$
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