Question

If x${}_1$, x${}_2$, . . . . x${}_{20}$ are in H.P and x${}_1$, 2, x${}_{20}$ are in G.P., then$\sum _{r=1}^{19}{x}_1{x}_{r+1}=$

• A. 76
• B. 80
• C. 84
• D. none of these

Answer 1

The correct option is A

76

Clearly,$\frac{1}{x_1},\frac{1}{x_2},....\frac{1}{x_{20}}$will be in A.P. Hence,

$\frac{1}{x_2}-\frac{1}{x_1}=\frac{1}{x_3}-\frac{1}{x_2}=....=\frac{1}{x_{r+1}}-\frac{1}{x_r}=....=\lambda \left(say\right)$

$\Rightarrow \frac{x_r-x_{r+1}}{x_r.x_{r+1}}=\lambda$

$\Rightarrow x_r.x_{r+1}=-\frac{1}{\lambda }(x_{r+1}-x_r)$

$\Rightarrow \sum _{r=1}^{19}{x}_r.x_{r+1}=-\frac{1}{\lambda }\sum _{r=1}^{19}(x_{r+1}-x_r)$

$=-\frac{1}{\lambda }(x_{20}-x_1)$

Now,

$\frac{1}{x_{20}}=\frac{1}{x_1}+19\lambda$

$\Rightarrow \frac{x_1-x_{20}}{x_1{x}_{20}}=19\lambda$

$\Rightarrow \sum _{r=1}^{19}{x}_r{x}_{r+1}=19x_1{x}_{20}=19\times 4=76$

$(\because x_1.2x_{20}$are in G.P., then $x_1{x}_{20}=4)$