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rtiit (436)

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Sorry guyz edit:

If a1, a2, a3,......... are in A.P. such that a1+a5+a10+a15+a20+a24=225
then a1+a2+a3+..............+a24=???

plz give detailed solution 

a15 is der i missed d term while typin
    

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rtiit (436)

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ne1 plz
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ankitagg (330)

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since it is an A.P with let common difference d;
a1+(a1+4d)+(a1+9d)+(a1+14d)+(a1+19d)+(a1+23d)=225
6a1+69d=225
2a1+23d=75   -(1)

sum of 24 terms
S=24/2[2(a1)+(24-1)d]
  =12[2a1+23d]
  =12(75)   [from 1]
  =900.
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computer001 (1849)

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where did a1 + 14d come from he hasnt said a15 in the sum

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rtiit (436)

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thnx
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akhil_o (2709)

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edit

" Always remember money isn't everything but make sure you have made a lot of it before talking such nonsense!"
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iitjee08aspirant (284)

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hay ankit u have taken one extra term....
a1+(a1+4d)+(a1+9d)+(a1+19d)+(a1+23d)=225
solving ul get
a1+11d=45
sum=24/2 [2a1+(24-1)d]
solving thin u shud get the ans

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sandeepramesh (1247)

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the problem is you dont know d so the answer cant be got :)
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rtiit (436)

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hey bt how did he get the rite ans???
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rtiit (436)

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mohit how can u further solve it??
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computer001 (1849)

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exactly rohit!!
but tht fellow who  1st posted the ans took the term a15 also...
spot nv i think...u agree??

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sandeepramesh (1247)

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 bcos you might have missed the term a15 in the quesand he included it
 
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kishore.subramanian.b (632)

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I think the question is a1+a5+a10+a15+a20+a24=225
Then.....
 
In an AP  an+a1=an-1+a2=........
Hence a1+a24=a2+a23=...........
 
3(a1+a24)=225
 a1+a24=75
 
Hence 24/2 (a1+a24) =12 (75) = 900
 
rate me if correct.........
 
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