Question

If $a,a_1,a_2.............a_{10},b$ are in A.P. and $a,g_1,g_2......g_{10},b$ are in G.P. and $h$ is the H.M between $a$ and $b$, then $\frac{a_1+a_2+.........+a_{10}}{g_1{g}_{10}}+\frac{a_2+a_3+.........+a_9}{g_2{g}_9}+...........+\frac{a_5+a_6}{g_5{g}_6}$ is

• A. $\frac{10}{h}$
• B. $\frac{15}{h}$
• C. $\frac{30}{h}$
• D. $\frac{5}{h}$

Answer 1

Answer: (C)

$\frac{30}{h}$

Given $a,a_1,a_2.............a_{10},b$ are in A.P.
Total number of terms in the AP are 12.
Let $d$ be the common difference.
$b=a+11d$
$\Rightarrow d=\frac{b-a}{11}$
$a_1=a+d=a+\frac{b-a}{11}=\frac{10a+b}{11}$
$a_2=a+2d=a+2\frac{(b-a)}{11}=\frac{9a+2b}{11}$
$a_9=a+9d=a+9\frac{(b-a)}{11}=\frac{9a+b}{11}$
$a_{10}=a+10d=a+10\frac{(b-a)}{11}=\frac{a+10b}{11}$
So, $a_1+a_{10}=a_2+a_9=..........=a+b$ .....(1)
Also,given $a,g_1,g_2......g_{10},b$ are in GP
Let $r$ be the common ratio
$b=a{r}^{11}$
$\Rightarrow r={\left(\frac{b}{a}\right)}^{1/11}$
Now, $g_1=ar=a{\left(\frac{b}{a}\right)}^{1/11}=a^{10/11}{b}^{1/11}$
$g_2=a{r}^2=a{\left(\frac{b}{a}\right)}^{2/11}=a^{9/11}{b}^{2/11}$
$g_9=a{r}^9=a{\left(\frac{b}{a}\right)}^{9/11}=a^{2/11}{b}^{9/11}$
$g_{10}=a{r}^{10}=a{\left(\frac{b}{a}\right)}^{10/11}=a^{1/11}{b}^{10/11}$
So, $g_1{g}_{10}=g_2{g}_9....=ab$ ....(2)
Consider,
$\frac{a_1+a_2+.........+a_{10}}{g_1{g}_{10}}+\frac{a_2+a_3+.........+a_9}{g_2{g}_9}+...........+\frac{a_5+a_6}{g_5{g}_6}$
$=\frac{5(a+b)+4(a+b)+3(a+b)+2(a+b)+(a+b)}{ab}$
$=15\frac{(a+b)}{ab}$
$=\frac{30}{h}(\therefore h=\frac{2ab}{a+b})$