Sequence And Series Question 249
Question: If the first and $ {{(2n-1)}^{th}} $ terms of an A.P., G.P. and H.P. are equal and their $ n^{th} $ terms are respectively $ a,\ b $ and $ c $ , then
[IIT 1985, 88]
Options:
A) $ a\ge b\ge c $
B) $ a+c=b $
C) $ ac-b^{2}=0 $
D) (a) and (c) both
Show Answer
Answer:
Correct Answer: D
Solution:
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 ) $ ?..(i) Again for G.P. : $ \beta =\alpha .{r^{2n-2}}\Rightarrow r={{( \frac{\beta }{\alpha } )}^{\frac{1}{2n-2}}} $
$ \therefore $ $ n^{th} $ term $ =b=\alpha {r^{n-1}}=\alpha {{( \frac{\beta }{\alpha } )}^{\frac{n-1}{2n-2}}}=\alpha {{( \frac{\beta }{\alpha } )}^{\frac{1}{2}}} $ or $ b={{(\alpha \beta )}^{1/2}}=\sqrt{\alpha \beta } $ ?..(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 } $ ?..(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 $
$ \therefore $ $ a\ge b\ge c $ ?..(iv) Now we try for : $ ac=\frac{\alpha +\beta }{2}.\frac{2\alpha \beta }{\alpha +\beta }=\alpha \beta =b^{2} $
$ \therefore $ $ ac-b^{2}=0 $ ?..(v) Obviously it can be seen that $ a+c\ne b $ ?..(vi) Hence (a) and (b) both hold good.
BETA
