Question

If a, b,c ,
d are in GP, then Prove that
a²-b² , b²-c² , c²-d² are also in G.P.
​

Answer 1

Step-by-step explanation:

a, b, c, d are in GP

Let the common ratio be r

1st term = a

2nd term = ar = b

3rd term = $a {r}^{2} = c$

4th term = $a {r}^{3} = d$

${a}^{2} - {b}^{2} = {a}^{2} - {(ar)}^{2} = {a}^{2} (1 - {r}^{2} )$

${b}^{2} - {c}^{2} = {(ar)}^{2} - { (a{r}^{2}) }^{2} = {a}^{2} {r}^{2} - {a}^{2} {r}^{4} = {a}^{2} {r}^{2} (1 - {r}^{2} )$

${c}^{2} - {d}^{2} = {(a {r}^{2}) }^{2} - {(a {r}^{3}) }^{2} = {a}^{2} {r}^{4} - {a}^{2} {r}^{6} = {a}^{2} {r}^{4} (1 - {r}^{2} )$

From above

${a}^{2} - {b}^{2} = {a}^{2} (1 - {r}^{2} )$

${b}^{2} - {c}^{2} = {a}^{2} (1 - {r}^{2} ) \times {r}^{2}$

${c}^{2} - {d}^{2} = {a}^{2} (1 - {r}^{2} ) \times {r}^{4}$

Therefore, ${a}^{2} - {b}^{2}, \: {b}^{2} - {c}^{2}, \: {c}^{2} - {d}^{2}$ are in GP with common difference $\: {r}^{2}$