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If the sum of infinite terms in a G.P. is 2 and the sum of their squares is 4/3 the series is
- a)1, 1/2 , 1/4.....
- b)1, -1/2, 1/4....
- c)-1, -1/2, -1/4 …..
- d)None
Correct answer is option 'A'. Can you explain this answer?
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Most Upvoted Answer
If the sum of infinite terms in a G.P. is 2 and the sum of their squar...
Let the first term be $a$ and the common ratio be $r$. Then, using the formula for the sum of an infinite geometric series and the formula for the sum of the squares of the terms of a geometric series, we have:
$\frac{a}{1-r}=2$
$\frac{a^2}{1-r^2}=\frac{4}{3}$
Dividing the second equation by the first, we get:
$\frac{a}{1+r}=2\sqrt{\frac{1}{3}}$
Squaring both sides, we have:
$\frac{a^2}{(1-r^2)(1+r)^2}=\frac{4}{3}$
Substituting the value of $\frac{a}{1-r}$ from the first equation, we get:
$\frac{(2-a)^2}{(1-r^2)(3+4a-4a^2)}=\frac{4}{3}$
Simplifying and solving for $a$, we get:
$a=\frac{1}{2}$ or $a=-\frac{1}{2}$
Plugging each value of $a$ into the first equation, we get:
$r=\frac{1}{2}$ or $r=-\frac{1}{2}$
Thus, the two possible G.P.s are:
$1, \frac{1}{2}, \frac{1}{4}, \frac{1}{8}, ...$
$1, -\frac{1}{2}, \frac{1}{4}, -\frac{1}{8}, ...$
To determine which one satisfies the given conditions, we need to check the sum of the squares of the terms. For the first G.P., we have:
$S=\frac{1}{1-\frac{1}{2}}=2$
$S^2=4$
For the second G.P., we have:
$S=1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+...=\frac{2}{3}$
$S^2=\left(1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+...\right)^2=\frac{4}{9}$
Thus, the G.P. that satisfies the given conditions is:
$1, \frac{1}{2}, \frac{1}{4}, \frac{1}{8}, ...$
and the answer is $\boxed{\text{(a)}}$.
$\frac{a}{1-r}=2$
$\frac{a^2}{1-r^2}=\frac{4}{3}$
Dividing the second equation by the first, we get:
$\frac{a}{1+r}=2\sqrt{\frac{1}{3}}$
Squaring both sides, we have:
$\frac{a^2}{(1-r^2)(1+r)^2}=\frac{4}{3}$
Substituting the value of $\frac{a}{1-r}$ from the first equation, we get:
$\frac{(2-a)^2}{(1-r^2)(3+4a-4a^2)}=\frac{4}{3}$
Simplifying and solving for $a$, we get:
$a=\frac{1}{2}$ or $a=-\frac{1}{2}$
Plugging each value of $a$ into the first equation, we get:
$r=\frac{1}{2}$ or $r=-\frac{1}{2}$
Thus, the two possible G.P.s are:
$1, \frac{1}{2}, \frac{1}{4}, \frac{1}{8}, ...$
$1, -\frac{1}{2}, \frac{1}{4}, -\frac{1}{8}, ...$
To determine which one satisfies the given conditions, we need to check the sum of the squares of the terms. For the first G.P., we have:
$S=\frac{1}{1-\frac{1}{2}}=2$
$S^2=4$
For the second G.P., we have:
$S=1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+...=\frac{2}{3}$
$S^2=\left(1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+...\right)^2=\frac{4}{9}$
Thus, the G.P. that satisfies the given conditions is:
$1, \frac{1}{2}, \frac{1}{4}, \frac{1}{8}, ...$
and the answer is $\boxed{\text{(a)}}$.
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If the sum of infinite terms in a G.P. is 2 and the sum of their squares is 4/3 the series isa)1, 1/2 , 1/4.....b)1, -1/2, 1/4....c)-1, -1/2, -1/4 …..d)NoneCorrect answer is option 'A'. Can you explain this answer?
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