Question

If first and $(2n-1{)}^{th}$ terms on an A.P,G.P and H.P are equal and their $n^{th}$ terms are $a,b,c$ respectively, then

• A. $a=b=c$
• B. $a+c=b$
• C. $a>b>c$
• D. $ac-b^2=0$

Answer 1

Answer: (C), (D)

The correct options are
C $a>b>c$
D $ac-b^2=0$

Given first and $(2n-1{)}^{th}$ term of an AP, GP and HP are same

AP: $a_1,a_2,a_3,a_4,....C.D.=d$

GP: $a,ar,a{r}^2,a{r}^3,.....C.D.=r$
HP: $\frac{1}{H_1},\frac{1}{H_2},\frac{1}{H_3},\frac{1}{H_4},....C.D.=d^{\prime }$

$a_1=a=\frac{1}{H_1}$

$a_{2n-1}=a_1+(2n-2)d=a{r}^{2n-2}=\frac{1}{H_1+(2n-2)d^{\prime }}a=a+(n-1)db=a{r}^{n-1}c=\frac{1}{H_1+(n-1)d^{\prime }}$

$a$ is the arithmetic mean of the first and $(2n-1{)}^{th}$ term.

$b$ is the geometric mean of the first and $(2n-1{)}^{th}$ term.

$c$ is the harmonic mean of the first and $(2n-1{)}^{th}$ term.

Hence, $a>b>c$

and $a\times c=b^{2}ac-b^2=0$