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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.
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Note by Aneesh Kundu 6 years, 2 months ago
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In an AP if a1+a5+a10+a15+a25=300, find sum up to 24 terms?
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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
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Top NewestIn an AP if a1+a5+a10+a15+a25=300, find sum up to 24 terms?
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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
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(Y- x)/z
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Why?
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d=nzy−x
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Hi Sudeep, Any tips for studying coordinate geometry for JEE?
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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!
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See my comment in the discussion.
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I think i got it correct.
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Well,what is it? How many did you solve in total?
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4
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I don't know whether my answer is right or not
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