Question-22090

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,.........

$$\left(\mathrm{i}\right)\:\mathrm{3},\:\:\mathrm{15},\:\:\:\mathrm{75},\:\:\:\mathrm{375},……. \\ $$$$\left(\mathrm{ii}\right)\mathrm{3},\:\:\mathrm{15},\:\:\:\mathrm{27},\:\:\:\mathrm{39},……… \\ $$

Commented by tawa tawa last updated on 11/Oct/17

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

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)

$${ATQ} \\ $$$${t}_{\mathrm{2}} ^{\mathrm{2}} ={t}_{\mathrm{1}} .{t}_{\mathrm{7}} \\ $$$$\left({a}+{d}\right)^{\mathrm{2}} ={a}\left({a}+\mathrm{6}{d}\right) \\ $$$${a}^{\mathrm{2}} +{d}^{\mathrm{2}} +\mathrm{2}{ad}={a}^{\mathrm{2}} +\mathrm{6}{ad} \\ $$$${d}^{\mathrm{2}} −\mathrm{4}{ad}=\mathrm{0} \\ $$$${d}\left({d}−\mathrm{4}{a}\right)=\mathrm{0} \\ $$$${d}=\mathrm{4}{a}\:\:−−\left(\mathrm{1}\right) \\ $$$${r}=\frac{{t}_{\mathrm{2}} }{{t}_{\mathrm{1}} }=\frac{{a}+{d}}{{a}}=\frac{\mathrm{5}{a}}{{a}}=\mathrm{5} \\ $$$${Let}\:{product}\:{of}\:{n}\:{terms}\:{of}\:{G}.{P}.=\mathrm{3375} \\ $$$$\Rightarrow{A}.{Ar}.{Ar}^{\mathrm{2}} ….{Ar}^{{n}−\mathrm{1}} =\mathrm{3375} \\ $$$$\Rightarrow{A}^{{n}} .{r}^{\frac{{n}\left({n}−\mathrm{1}\right)}{\mathrm{2}}} =\mathrm{3375} \\ $$$$\Rightarrow{A}^{{n}} .\mathrm{5}^{\frac{{n}\left({n}−\mathrm{1}\right)}{\mathrm{2}}} =\mathrm{3}^{\mathrm{3}} \mathrm{5}^{\mathrm{3}} \\ $$$$\Rightarrow{n}=\mathrm{3},\:{A}=\mathrm{3} \\ $$$$\therefore\:{the}\:{Geometric}\:{series}\:{is} \\ $$$$\mathrm{3}+\mathrm{15}+\mathrm{75}\:\:{Ans}.\left({i}\right) \\ $$$${Now},\:{t}_{\mathrm{2}} ={a}+{d} \\ $$$$\Rightarrow{Ar}={a}+\mathrm{4}{a} \\ $$$$\Rightarrow\mathrm{15}=\mathrm{5}{a} \\ $$$$\Rightarrow{a}=\mathrm{3}\:\&{d}=\mathrm{12} \\ $$$$\therefore\:{using}\:{formula}\:{S}_{{n}} =\frac{{n}}{\mathrm{2}}\left\{\mathrm{2}{a}+\left({n}−\mathrm{1}\right){d}\right\} \\ $$$${S}_{{n}} =\frac{{n}}{\mathrm{2}}\left\{\mathrm{6}+\left({n}−\mathrm{1}\right).\mathrm{12}\right\} \\ $$$$\:{S}_{{n}} =\frac{{n}}{\mathrm{2}}\left(\mathrm{12}{n}−\mathrm{6}\right)\:{Ans}.\left({ii}\right) \\ $$$$ \\ $$

Commented by tawa tawa last updated on 11/Oct/17

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

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