Question

if a b c d e are five numbers in which the first three are in AP the last three are in HP. If three numbers in the middle are in GP then the numbers in the odd places are in? a-ap b-gp c-hp d-none ​

Answer 1

Answer: b-GP

Step-by-step explanation:

1) We have,

Five numbers a,b,c,d &e.

The first three are in AP.

=>a,b,c are in AP

=>2b=a+c

2) The last three numbers are in HP.

=>c,d,e are in HP.

=>1/c,1/d,1/e are in AP.

$=>2\frac{1}{d}=\frac{1}{c}+\frac{1}{e}\\ \\=>\frac{2}{d}=\frac{c+e}{ce}\\ \\=>d=\frac{2ce}{c+e}$

3) Middle three numbers are in GP.

=>b,c,d are in GP.

$=>c^2=bd$

4) Now, we need to check the numbers in the odd places.

=> Check: a,c,e

We will find relationship between a,c, &e.

5) That is,Starting from

$2b=a+c \\ \\=>a=2b-c$

$=> a=2*\frac{c^2}{d}-c \\ \\=>a=2*\frac{c^2}{\frac{2ce}{c+e}}-c$

$=>a=\frac{c(c+e)}{e}-c=\frac{c^}{e}+c-c=\frac{c^2}{e}$

$=>ae=c^2$

Hence, a,c,e are in GP.=> The numbers in the odd places are in GP.