koli123able
The sum of first 14 terms of 1,1.4,1.8,.... is equal to the sum of first n terms of 5.6,5.8,6.0,.... , find the value of n



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sauravshakya
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1,1.4,1.8,....
This is in AP with first term 1 and common difference 0.4

sauravshakya
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Use sum formula of AP

amoodarya
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14/2(a1+a14)=n/2(5.6+5.6+(n1)*0.2)

amoodarya
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14/2(1+1+13(.4))=n/2(5.6+5.6+(n1)*0.2)

koli123able
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i did the first A.P and its coming 50.4

sauravshakya
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S14 =14/2 {2*1 + (141)0.4} = 50.4

koli123able
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i am stuck here

sauravshakya
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5.6,5.8,6.0,....
This is in AP with common difference 0.2 andfirt term 5.6

koli123able
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\[50.4=\frac{ n }{ 2 }[(2*5.6)+(n1)*0.2]\]

koli123able
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stuck here

sauravshakya
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Sn=50.4
n/2 { 2*5.6 + (n1)0.2}=50.4
n{11  0.2n}=100.8

koli123able
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2*5.6 its 11.2

sauravshakya
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11n 0.2n^2 =100.8
11n 2/10 n^2 =100.8
110n 2n^2 =1008

sauravshakya
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11.20.2=11

koli123able
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oh

sauravshakya
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110n 2n^2 =1008
2n^2110n+1008=0
n^255n+504=0

koli123able
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how did became 110n 2n^2 =1008

sauravshakya
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dw:1359028396466:dw

koli123able
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ok

sauravshakya
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Now,n^255n+504=0
CAN u solve for n

koli123able
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n= 11.61 and n= 43.38

koli123able
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now what?

koli123able
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The Answer should come n=8...i checked the answers

sauravshakya
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n=8 is surely not the correct solution

koli123able
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the book says that answer is n=8

koli123able
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how will it come 8

sauravshakya
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Sorry I did a algebra mistake

sauravshakya
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n=8 is correct

koli123able
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how????? plz plz tell me

sauravshakya
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Sn=50.4
n/2 { 2*5.6 + (n1)0.2}=50.4
n/2 {11+0.2n}=50.4
n{11+0.2n}=100.8
11n+0,2n^2=100.8
11n +2/10 n^2 =100.8
110n+2n^2=1008
55n+n^2=504
n^2+55n504=0
n^2+(638)n504=0
n^2 +63n 8n504=0
n(n+63)8(n+63)=0
(n8)(n+63)=0

sauravshakya
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Now, either n8=0
n=8
OR n+63=0
n=63
BUT n cannot be negative
So, n=8

sauravshakya
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got it?

koli123able
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thanks

sauravshakya
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welcome