Forgot password? New user? Sign up
Existing user? Log in
Let \(a_1,a_2,\ldots ,a_{2n}\) be an arithematic progression of positive real numbers with common difference \(d\) . Let
{∑i=1na2i−12=x ∑i=1na2i2=y an+an+1=z\large\left\{\begin{array}{l}\displaystyle\sum^n_{i=1} a^2_{2i-1}=x\\\ \displaystyle\sum^n_{i=1} a^2_{2i}=y\\\ a_n+a_{n+1}=z\end{array}\right.⎩⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎧i=1∑na2i−12=x i=1∑na2i2=y an+an+1=z
Express ddd in terms of x,y,z,nx, y, z, nx,y,z,n.
You can find rest of the problems here
You can find the solutions here
Note by Aneesh Kundu 6 years, 2 months ago
Easy Math Editor
This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
When posting on Brilliant:
*italics*
_italics_
**bold**
__bold__
- bulleted- list
1. numbered2. list
paragraph 1paragraph 2
paragraph 1
paragraph 2
[example link](https://brilliant.org)
> This is a quote
This is a quote
# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"
\(
\)
\[
\]
2 \times 3
2^{34}
a_{i-1}
\frac{2}{3}
\sqrt{2}
\sum_{i=1}^3
\sin \theta
\boxed{123}
Sort by:
In an AP if a1+a5+a10+a15+a25=300, find sum up to 24 terms?
Log in to reply
Got the same question in Tamil Nadu's RMO. My approach was as follows:
y−x=(a2n2−a12)+(a2n−22−a32)+...+(a22−a2n−12)y-x=(a^{ 2 }_{ 2n }-a^{ 2 }_{ 1 })+(a^{ 2 }_{ 2n-2 }-a^{ 2 }_{ 3 })+...+(a^{ 2 }_{ 2 }-a^{ 2 }_{ 2n-1 })y−x=(a2n2−a12)+(a2n−22−a32)+...+(a22−a2n−12)
It is a property of an AP that the sum of equidistant terms from the middle term(s) is a constant. Note that ana_nan and an+1a_{n+1}an+1 are the middle terms of this AP. Hence, their sum zzz is constant for all equidistant terms.
y−x=(a2n+a1)(a2n−a1)+(a2n−2+a3)(a2n−2−a3)+...+(a2+a2n−1)(a2−a2n−1)y-x=(a_{ 2n }+a_{ 1 })(a_{ 2n }-a_{ 1 })+(a_{ 2n-2 }+a_{ 3 })(a_{ 2n-2 }-a_{ 3 })+...+(a_{ 2 }+a_{ 2n-1 })(a_{ 2 }-a_{ 2n-1 })y−x=(a2n+a1)(a2n−a1)+(a2n−2+a3)(a2n−2−a3)+...+(a2+a2n−1)(a2−a2n−1)
y−x=z(a2n−a1)+z(a2n−2−a3)+...+z(a2−a2n−1)y-x=z(a_{ 2n }-a_{ 1 }) + z(a_{ 2n-2 }-a_{ 3 }) + ... + z(a_{ 2 }-a_{ 2n-1 })y−x=z(a2n−a1)+z(a2n−2−a3)+...+z(a2−a2n−1)
y−x=z(a2n−a1+a2n−2−a3+a2−a2n−1)y-x=z(a_{ 2n }-a_{ 1 } + a_{ 2n-2 }-a_{ 3 } +a_{ 2 }-a_{ 2n-1 })y−x=z(a2n−a1+a2n−2−a3+a2−a2n−1)
y−x=z((a2n−a2n−1)+(a2n−2−a2n−3)+...+(a2−a1))y-x=z ((a_{2n}-a_{2n-1}) + (a_{2n-2} - a_{2n-3}) + ... + (a_2-a_1))y−x=z((a2n−a2n−1)+(a2n−2−a2n−3)+...+(a2−a1))
The above terms are the difference of two consecutive terms, ddd, and there are nnn such terms:
y−x=zndy-x=zndy−x=znd
d=y−xznd=\dfrac{y-x}{zn}d=zny−x
(Y- x)/z
Why?
d=y−xnz \displaystyle d = \frac{y-x}{nz} d=nzy−x
Hi Sudeep, Any tips for studying coordinate geometry for JEE?
got the same answer yesterday in RMO
its same as in RMO Karnataka Region!
I got a equation with x,y,z,n,d and d^2.. Couldn't go further!
Did u try using quadratic formula after that?
No! I didn't. May be the equation was wrong, Sudeep Salgia's solution looks right!
@Ranjana Kasangeri – See my comment in the discussion.
I think i got it correct.
Well,what is it? How many did you solve in total?
@Ranjana Kasangeri – 4
@Aneesh Kundu – I don't know whether my answer is right or not
Problem Loading...
Note Loading...
Set Loading...
Easy Math Editor
This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
When posting on Brilliant:
*italics*
or_italics_
**bold**
or__bold__
paragraph 1
paragraph 2
[example link](https://brilliant.org)
> This is a quote
\(
...\)
or\[
...\]
to ensure proper formatting.2 \times 3
2^{34}
a_{i-1}
\frac{2}{3}
\sqrt{2}
\sum_{i=1}^3
\sin \theta
\boxed{123}
Comments
Sort by:
Top NewestIn an AP if a1+a5+a10+a15+a25=300, find sum up to 24 terms?
Log in to reply
Got the same question in Tamil Nadu's RMO. My approach was as follows:
y−x=(a2n2−a12)+(a2n−22−a32)+...+(a22−a2n−12)
It is a property of an AP that the sum of equidistant terms from the middle term(s) is a constant. Note that an and an+1 are the middle terms of this AP. Hence, their sum z is constant for all equidistant terms.
y−x=(a2n+a1)(a2n−a1)+(a2n−2+a3)(a2n−2−a3)+...+(a2+a2n−1)(a2−a2n−1)
y−x=z(a2n−a1)+z(a2n−2−a3)+...+z(a2−a2n−1)
y−x=z(a2n−a1+a2n−2−a3+a2−a2n−1)
y−x=z((a2n−a2n−1)+(a2n−2−a2n−3)+...+(a2−a1))
The above terms are the difference of two consecutive terms, d, and there are n such terms:
y−x=znd
d=zny−x
Log in to reply
(Y- x)/z
Log in to reply
Why?
Log in to reply
d=nzy−x
Log in to reply
Hi Sudeep, Any tips for studying coordinate geometry for JEE?
Log in to reply
got the same answer yesterday in RMO
Log in to reply
its same as in RMO Karnataka Region!
I got a equation with x,y,z,n,d and d^2.. Couldn't go further!
Log in to reply
Did u try using quadratic formula after that?
Log in to reply
No! I didn't. May be the equation was wrong, Sudeep Salgia's solution looks right!
Log in to reply
Log in to reply
I think i got it correct.
Log in to reply
Well,what is it? How many did you solve in total?
Log in to reply
Log in to reply
Log in to reply