જો A.P., G.P. અને H.P. પ્રથમ અને {(2n - 1)^{t | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(d) Let $\alpha ,\beta $ be the first and ${(2n - 1)^{th}}$ terms of the $A.P.$,

the $G.P. $and the $H.P.$ respectively. Then we have

For $A.P.$

Vedclass · Accepted Answer

$(a)$ અને $(c)$ બંને

Vedclass · Answer

(d) Let $\alpha ,\beta $ be the first and ${(2n - 1)^{th}}$ terms of the $A.P.$,

the $G.P. $and the $H.P.$ respectively. Then we have

For $A.P.$ : $\beta = \alpha + (2n - 2)d \Rightarrow d = \frac{{\beta - \alpha }}{{2n - 2}}$

${n^{th}}$term $ = a = \alpha + (n - 1)d = \frac{1}{2}(\alpha + \beta )$&hellip;..(i)

Again for $G.P.$ : $\beta = \alpha .{r^{2n - 2}} \Rightarrow r = {\left( {\frac{\beta }{\alpha }} ight)^{\frac{1}{{2n - 2}}}}$

$	herefore $${n^{th}}$ term $ = b = \alpha {r^{n - 1}} = \alpha {\left( {\frac{\beta }{\alpha }} ight)^{\frac{{n - 1}}{{2n - 2}}}} = \alpha {\left( {\frac{\beta }{\alpha }} ight)^{\frac{1}{2}}}$

or $b = {(\alpha \beta )^{1/2}} = \sqrt {\alpha \beta } $&hellip;..(ii)

Again for $H.P. $: $\frac{1}{\beta } = \frac{1}{\alpha } + (2n - 2)d'$

$ \Rightarrow $$\frac{1}{c} = \frac{1}{\alpha } + (n - 1)d' = \frac{1}{\alpha } + \frac{{\alpha - \beta }}{{2\alpha \beta }} = \frac{{\alpha + \beta }}{{2\alpha \beta }}$

$ \Rightarrow $$c = \frac{{2\alpha \beta }}{{\alpha + \beta }}$ &hellip;..(iii)

Now, more than one of the alternative answers may be correct. We try for $(a)$ :

$a - b = \frac{{\alpha + \beta }}{2} - \sqrt {\alpha \beta } = \frac{1}{2}{(\sqrt \alpha - \sqrt \beta )^2} \ge 0$

$\Rightarrow a \ge b$

$b - c = \sqrt {\alpha \beta } - \frac{{2\alpha \beta }}{{\alpha + \beta }} = \frac{{\sqrt {\alpha \beta } }}{{\alpha + \beta }}(\alpha + \beta - 2\sqrt {\alpha \beta } )$

$ = \frac{{\sqrt {\alpha \beta } }}{{(\alpha + \beta )}}{(\sqrt \alpha - \sqrt \beta )^2} \ge 0 $

$\Rightarrow b \ge c$

$	herefore $$a \ge b \ge c$&hellip;..(iv)

Now we try for $(c)$ : $ac = \frac{{\alpha + \beta }}{2}.\frac{{2\alpha \beta }}{{\alpha + \beta }} = \alpha \beta = {b^2}$

$	herefore $$ac - {b^2} = 0$&hellip;..(v)

Obviously it can be seen that $a + c 
e b$&hellip;..(vi)

Hence $(a)$ and $(c)$ both hold good.

જો $A.P., G.P.$ અને $H.P.$ પ્રથમ અને ${(2n - 1)^{th}}$ પદના સમાન હોય અને તેમના ${n^{th}}$ પદો અનુક્રમે $a,b$ અને $c$ હોય તો

Similar Questions

જો $A, G, H$ આપેલી બે સંખ્યાઓ વચ્ચેના અનુક્રમે સમાંતર મધ્યક, સમગુણોત્તર મધ્યક અને સ્વરીત મધ્યક હોય, તો = …..

જો $a, b, c$ સમાંતર શ્રેણીમાં તથા સમગુણોત્તર શ્રેણીમાં હોય, તો......