Question
State whether statement are True or False.
The sum or difference of two G.P.s, is again a G.P.
The sum or difference of two G.P.s, is again a G.P.
✓
Answer
False.
Solution:
Let two G.P.s are $\text{a},\text{ar}_1,\text{ar}_1^2,\text{ar}_1^3....;$ and $\text{b},\text{br}_2,\text{br}_2^2,\text{br}_2^3,....$
Now, sum of two G.P.s is $\text{a}+\text{b},(\text{ar}_1+\text{br}_2),\big(\text{ar}_1^2+\text{br}_2^2\big),....$
Clearly,
$\frac{\text{ar}_1+\text{br}_2}{\text{a + b}}\neq\frac{\text{ar}_1^2+\text{br}_2^2}{\text{ar}_1+\text{br}_2}$
Similarly, for difference of two G.P.s, we get
$\frac{\text{ar}_1+\text{br}_2}{\text{a + b}}\neq\frac{\text{ar}_1^2+\text{br}_2^2}{\text{ar}_1+\text{br}_2}$
So, the sum or difference of two G.P.s is not a G.P.
Solution:
Let two G.P.s are $\text{a},\text{ar}_1,\text{ar}_1^2,\text{ar}_1^3....;$ and $\text{b},\text{br}_2,\text{br}_2^2,\text{br}_2^3,....$
Now, sum of two G.P.s is $\text{a}+\text{b},(\text{ar}_1+\text{br}_2),\big(\text{ar}_1^2+\text{br}_2^2\big),....$
Clearly,
$\frac{\text{ar}_1+\text{br}_2}{\text{a + b}}\neq\frac{\text{ar}_1^2+\text{br}_2^2}{\text{ar}_1+\text{br}_2}$
Similarly, for difference of two G.P.s, we get
$\frac{\text{ar}_1+\text{br}_2}{\text{a + b}}\neq\frac{\text{ar}_1^2+\text{br}_2^2}{\text{ar}_1+\text{br}_2}$
So, the sum or difference of two G.P.s is not a G.P.
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