Question

Let $b_1,b_2,b_3,..b_{10}$ be in A.P; $g_1,g_2,g_3,...g_{10}$ be in H.P. If $b_1=g_1=1,b_{10}=g_{10}=3$ then the value of $b_4{g}_7$:

• A. $2$
• B. $3$
• C. $6$
• D. $7$

Answer 1

The correct option is C $6$

Given: $b_1,b_2,b_3,..b_{10}$ are in A.P

And $b_{10}=3$(given)

$\Rightarrow b_1+9d=3$

$\Rightarrow 2+9d=3$ (Given $b_1=2$)

$\Rightarrow 9d=3-2=1$

$\therefore d=\frac{1}{9}$

Now, $b_4=b_1+3d=2+3\times \frac{1}{9}=\frac{7}{3}$

$g_1,g_2,g_3,...g_{10}$ are in H.P

$\Rightarrow \frac{1}{g_1},\frac{1}{g_2},\frac{1}{g_3},..\frac{1}{g_{10}}$ are in A.P

$\Rightarrow \frac{1}{g_1}=\frac{1}{2},..\frac{1}{g_{10}}=\frac{1}{g_1}+9d$

$9d=\frac{1}{g_{10}}-\frac{1}{g_1}=\frac{1}{3}-\frac{1}{2}=-\frac{1}{6}$

$\Rightarrow 9d=-\frac{1}{6}$

$\therefore d=-\frac{1}{54}$

Again $\frac{1}{g_7}=\frac{1}{g_1}+6d=\frac{1}{2}+6\times \frac{1}{54}$

$\frac{1}{g_7}=\frac{1}{2}-\frac{1}{9}$ (on simplification)

$=\frac{9-2}{18}=\frac{7}{18}$

$\therefore g_7=\frac{18}{7}$

Hence, $b_4{g}_7=\frac{7}{3}\times \frac{18}{7}=6$