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Finding the sum of an Infinite GP given $a_1 = 1$ and $4a_2 + 5a_3 $ has least value
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I solved a question in a certain way that didn't seem right. I would like to know if there is any other way to do the question or if there is any fault in my logic for the question.
The question says to find the sum of a $infty$ GP such that $a_1 = 1$ and $4a_2 + 5a_3$ has the least value in GP.
My approach:
I solved by first assuming that since $4a_2 + 5a_3$ has the least value thus we can say, $$5r^2 + 4r lt 1$$
So, $$ 5r^2 + 5r -r -1 lt 0$$
Which gives us,
$$(5r-1)(r+1) lt0$$
So, $$r in Bigl(-1,frac15Bigl)$$
But, $$5r^2 + 4r lt 1 iff r lt 0$$
So $$r in Bigl(-1,0Bigl)$$
Now we know, $$sum ^infty_r=1 a_r = fraca1-r$$
So, since $$rlt 0$$ we get $$sum ^infty_r=1 a_r = fraca1+r; forall ;r in Bigl(-1,0Bigl)$$ This implies, since $$a = 1$$ That, $$sum ^infty_r=1 a_r = fraca1+r lt 1$$
Only one of the options matches this, and so that is how I arrived at the answer.
Please tell me if there is any flaw in the solution and also how to reach at a precise value.
The answer given is $frac57$ This matches my solution but I still felt like I could do with a second opinion.
sequences-and-series
edited Jul 27 at 4:45
Piyush Divyanakar
3,258122
asked Jul 27 at 4:42
Prakhar Nagpal
428315
add a comment |Â
up vote
0
down vote
favorite
I solved a question in a certain way that didn't seem right. I would like to know if there is any other way to do the question or if there is any fault in my logic for the question.
The question says to find the sum of a $infty$ GP such that $a_1 = 1$ and $4a_2 + 5a_3$ has the least value in GP.
My approach:
I solved by first assuming that since $4a_2 + 5a_3$ has the least value thus we can say, $$5r^2 + 4r lt 1$$
So, $$ 5r^2 + 5r -r -1 lt 0$$
Which gives us,
$$(5r-1)(r+1) lt0$$
So, $$r in Bigl(-1,frac15Bigl)$$
But, $$5r^2 + 4r lt 1 iff r lt 0$$
So $$r in Bigl(-1,0Bigl)$$
Now we know, $$sum ^infty_r=1 a_r = fraca1-r$$
So, since $$rlt 0$$ we get $$sum ^infty_r=1 a_r = fraca1+r; forall ;r in Bigl(-1,0Bigl)$$ This implies, since $$a = 1$$ That, $$sum ^infty_r=1 a_r = fraca1+r lt 1$$
Only one of the options matches this, and so that is how I arrived at the answer.
Please tell me if there is any flaw in the solution and also how to reach at a precise value.
The answer given is $frac57$ This matches my solution but I still felt like I could do with a second opinion.
sequences-and-series
edited Jul 27 at 4:45
Piyush Divyanakar
3,258122
asked Jul 27 at 4:42
Prakhar Nagpal
428315
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I solved a question in a certain way that didn't seem right. I would like to know if there is any other way to do the question or if there is any fault in my logic for the question.
The question says to find the sum of a $infty$ GP such that $a_1 = 1$ and $4a_2 + 5a_3$ has the least value in GP.
My approach:
I solved by first assuming that since $4a_2 + 5a_3$ has the least value thus we can say, $$5r^2 + 4r lt 1$$
So, $$ 5r^2 + 5r -r -1 lt 0$$
Which gives us,
$$(5r-1)(r+1) lt0$$
So, $$r in Bigl(-1,frac15Bigl)$$
But, $$5r^2 + 4r lt 1 iff r lt 0$$
So $$r in Bigl(-1,0Bigl)$$
Now we know, $$sum ^infty_r=1 a_r = fraca1-r$$
So, since $$rlt 0$$ we get $$sum ^infty_r=1 a_r = fraca1+r; forall ;r in Bigl(-1,0Bigl)$$ This implies, since $$a = 1$$ That, $$sum ^infty_r=1 a_r = fraca1+r lt 1$$
Only one of the options matches this, and so that is how I arrived at the answer.
Please tell me if there is any flaw in the solution and also how to reach at a precise value.
The answer given is $frac57$ This matches my solution but I still felt like I could do with a second opinion.
sequences-and-series
edited Jul 27 at 4:45
Piyush Divyanakar
3,258122
asked Jul 27 at 4:42
Prakhar Nagpal
428315
I solved a question in a certain way that didn't seem right. I would like to know if there is any other way to do the question or if there is any fault in my logic for the question.
The question says to find the sum of a $infty$ GP such that $a_1 = 1$ and $4a_2 + 5a_3$ has the least value in GP.
My approach:
I solved by first assuming that since $4a_2 + 5a_3$ has the least value thus we can say, $$5r^2 + 4r lt 1$$
So, $$ 5r^2 + 5r -r -1 lt 0$$
Which gives us,
$$(5r-1)(r+1) lt0$$
So, $$r in Bigl(-1,frac15Bigl)$$
But, $$5r^2 + 4r lt 1 iff r lt 0$$
So $$r in Bigl(-1,0Bigl)$$
Now we know, $$sum ^infty_r=1 a_r = fraca1-r$$
So, since $$rlt 0$$ we get $$sum ^infty_r=1 a_r = fraca1+r; forall ;r in Bigl(-1,0Bigl)$$ This implies, since $$a = 1$$ That, $$sum ^infty_r=1 a_r = fraca1+r lt 1$$
Only one of the options matches this, and so that is how I arrived at the answer.
Please tell me if there is any flaw in the solution and also how to reach at a precise value.
The answer given is $frac57$ This matches my solution but I still felt like I could do with a second opinion.
sequences-and-series
edited Jul 27 at 4:45
Piyush Divyanakar
3,258122
asked Jul 27 at 4:42
Prakhar Nagpal
428315
edited Jul 27 at 4:45
Piyush Divyanakar
3,258122
edited Jul 27 at 4:45
Piyush Divyanakar
3,258122
edited Jul 27 at 4:45
Piyush Divyanakar
3,258122
3,258122
asked Jul 27 at 4:42
Prakhar Nagpal
428315
asked Jul 27 at 4:42
Prakhar Nagpal
428315
asked Jul 27 at 4:42
Prakhar Nagpal
428315
428315
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
$$4a_2+5a_3=4r+5r^2$$
This has the least value when the derivative is $0$. $$frac ddr4r+5r^2=4+10r=0 \ implies r = frac-25$$
Hence sum of GP is $$frac11-frac-25=frac57$$
Alternatively if you don't know any calculus then note that for a quadratic function $f(x)=ax^2+bx+c$, minima is given by $-b/2a$. The midpoint of the roots of equation $f(x)=0$.
answered Jul 27 at 4:49
Piyush Divyanakar
3,258122
Can you tell me if there is any mistake in my solution?
– Prakhar Nagpal
Jul 27 at 6:01
Why do you think $5r^2+4r<1$ is true. There is no reason to believe this,
– Piyush Divyanakar
Jul 27 at 6:29
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
$$4a_2+5a_3=4r+5r^2$$
This has the least value when the derivative is $0$. $$frac ddr4r+5r^2=4+10r=0 \ implies r = frac-25$$
Hence sum of GP is $$frac11-frac-25=frac57$$
Alternatively if you don't know any calculus then note that for a quadratic function $f(x)=ax^2+bx+c$, minima is given by $-b/2a$. The midpoint of the roots of equation $f(x)=0$.
answered Jul 27 at 4:49
Piyush Divyanakar
3,258122
Can you tell me if there is any mistake in my solution?
– Prakhar Nagpal
Jul 27 at 6:01
Why do you think $5r^2+4r<1$ is true. There is no reason to believe this,
– Piyush Divyanakar
Jul 27 at 6:29
add a comment |Â
up vote
2
down vote
accepted
$$4a_2+5a_3=4r+5r^2$$
This has the least value when the derivative is $0$. $$frac ddr4r+5r^2=4+10r=0 \ implies r = frac-25$$
Hence sum of GP is $$frac11-frac-25=frac57$$
Alternatively if you don't know any calculus then note that for a quadratic function $f(x)=ax^2+bx+c$, minima is given by $-b/2a$. The midpoint of the roots of equation $f(x)=0$.
answered Jul 27 at 4:49
Piyush Divyanakar
3,258122
Can you tell me if there is any mistake in my solution?
– Prakhar Nagpal
Jul 27 at 6:01
Why do you think $5r^2+4r<1$ is true. There is no reason to believe this,
– Piyush Divyanakar
Jul 27 at 6:29
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
$$4a_2+5a_3=4r+5r^2$$
This has the least value when the derivative is $0$. $$frac ddr4r+5r^2=4+10r=0 \ implies r = frac-25$$
Hence sum of GP is $$frac11-frac-25=frac57$$
Alternatively if you don't know any calculus then note that for a quadratic function $f(x)=ax^2+bx+c$, minima is given by $-b/2a$. The midpoint of the roots of equation $f(x)=0$.
answered Jul 27 at 4:49
Piyush Divyanakar
3,258122
$$4a_2+5a_3=4r+5r^2$$
This has the least value when the derivative is $0$. $$frac ddr4r+5r^2=4+10r=0 \ implies r = frac-25$$
Hence sum of GP is $$frac11-frac-25=frac57$$
Alternatively if you don't know any calculus then note that for a quadratic function $f(x)=ax^2+bx+c$, minima is given by $-b/2a$. The midpoint of the roots of equation $f(x)=0$.
answered Jul 27 at 4:49
Piyush Divyanakar
3,258122
edited Jul 27 at 4:52
answered Jul 27 at 4:49
Piyush Divyanakar
3,258122
answered Jul 27 at 4:49
Piyush Divyanakar
3,258122
answered Jul 27 at 4:49
Piyush Divyanakar
3,258122
3,258122
Can you tell me if there is any mistake in my solution?
– Prakhar Nagpal
Jul 27 at 6:01
Why do you think $5r^2+4r<1$ is true. There is no reason to believe this,
– Piyush Divyanakar
Jul 27 at 6:29
add a comment |Â
Can you tell me if there is any mistake in my solution?
– Prakhar Nagpal
Jul 27 at 6:01
Why do you think $5r^2+4r<1$ is true. There is no reason to believe this,
– Piyush Divyanakar
Jul 27 at 6:29
Can you tell me if there is any mistake in my solution?
– Prakhar Nagpal
Jul 27 at 6:01
Can you tell me if there is any mistake in my solution?
– Prakhar Nagpal
Jul 27 at 6:01
Why do you think $5r^2+4r<1$ is true. There is no reason to believe this,
– Piyush Divyanakar
Jul 27 at 6:29
Why do you think $5r^2+4r<1$ is true. There is no reason to believe this,
– Piyush Divyanakar
Jul 27 at 6:29
add a comment |Â
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