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09-05-2005, 11:07 AM

Quote:
Originally Posted by catquery
ur answer is right but i did it in different way

if we take ath term as -1 then a times ath term will be -1*-1=1 similarly
if we take bth term as 1 then b times bth term will be 1*1=1
a times ath term=b times bth term i.e. 1 but
(a+b)th term will be = -1+1 = 0 as a = -1 and b = 1

tell me whether i am right or wrong
Seems to me like you are assuming too many things. There may be question which wil lead you in traps if you work out with such trivial exaples.

Quote:
Originally Posted by catquery
one more question
a number 20 is divided into four parts that are in AP such that the product of the first and fourth is to the product of the second and third is 2:3. Find the largest part.
(a) 12 (b) 4 (c) 8 (d) 9
//Writing convention : a2 = a^2 and 2a = 2*a
As there are four number in AP assume a-3d, a-d, a+d, a+3d
Sum = 20 => a=5
3*(a2 - 9d2) = 2*(a2-d2)
a2 = 25d2 => d=1, Largest number is a+3d = 8

Hint:
1) when you are given odd numbers of terms in AP assume a-2d, a-d , a, a +2d
2) when you are given even number of terms auusme a-5d, a-3d, a-d, a+d, a+3d, a+5d (rememer common difference is 2d)
   
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help - 13-05-2005, 04:05 PM

A geometric progression consists of 500 terms. Sum of the terms occupying the odd places is P1 and the sum of the terms occupying the even places is P2. Find the common ratio.

(a) p2/P1 (b) P1/P2 (c) P2+P1/P1 (d) P2+P1/P2
   
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13-05-2005, 04:27 PM

Quote:
Originally Posted by catquery
A geometric progression consists of 500 terms. Sum of the terms occupying the odd places is P1 and the sum of the terms occupying the even places is P2. Find the common ratio.

(a) p2/P1 (b) P1/P2 (c) P2+P1/P1 (d) P2+P1/P2

// Terminology r2 ~ r^2, series a, ar, ar2, ar3 ...

Sum of odd terms: a + ar2 + ar4 ... ar498 = P1
sum of odd terms: ar + ar3 + ar5 .... ar499 = P2 = r*P1

r = P2/P1 So Simple !!
   
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Simpler way
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Simpler way - 13-05-2005, 05:26 PM

Quote:
Originally Posted by Eccentric
// Terminology r2 ~ r^2, series a, ar, ar2, ar3 ...

Sum of odd terms: a + ar2 + ar4 ... ar498 = P1
sum of odd terms: ar + ar3 + ar5 .... ar499 = P2 = r*P1

r = P2/P1 So Simple !!
A better n simpler approach wud b to assume values for sumething like 4 nos in GP n chek
whats the common ratio....

assume for eg:nos are 1, 2, 4, 8...

Sum of even terms(P2)=10
Sum of od terms(P1)=5...

P2/P1 =10/5 ==2=r=common ratio
   
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13-05-2005, 05:35 PM

[quote=Eccentric]Seems to me like you are assuming too many things. There may be question which wil lead you in traps if you work out with such trivial exaples.





Well u r just require to assume a=1 (rather than -1), rest all follows automatically naa....
after asuming a=1 u get b=3 ...n the sum bcoms extremely easy to solve ....

Bu it all depends on the approach u r comfartable with...

keep working
   
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14-05-2005, 12:23 AM

ur answer is right and for me this approach is also easier but one thing u tell me that everytime we have to make assumption in numerical value or we can make assumption as eccentric did

one more problem can u solve
A number of saplings are lying at a place by the side of a straight road. These are to be planted in a straight line at a distance interval of 10 metres between two consecutive saplings. Mithilesh, the country`s greatest forester, can carry only one sapling at a time and has to move back to the original point to get the next sapling. In this manner he covers a total distance of 1.32 kms. How many saplings does he plant in the process.

(a) 15 (b) 14 (c) 13 (d) 12

Quote:
Originally Posted by Apprentice
A better n simpler approach wud b to assume values for sumething like 4 nos in GP n chek
whats the common ratio....

assume for eg:nos are 1, 2, 4, 8...

Sum of even terms(P2)=10
Sum of od terms(P1)=5...

P2/P1 =10/5 ==2=r=common ratio
   
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14-05-2005, 12:42 AM

Quote:
Originally Posted by catquery
ur answer is right and for me this approach is also easier but one thing u tell me that everytime we have to make assumption in numerical value or we can make assumption as eccentric did

one more problem can u solve
A number of saplings are lying at a place by the side of a straight road. These are to be planted in a straight line at a distance interval of 10 metres between two consecutive saplings. Mithilesh, the country`s greatest forester, can carry only one sapling at a time and has to move back to the original point to get the next sapling. In this manner he covers a total distance of 1.32 kms. How many saplings does he plant in the process.

(a) 15 (b) 14 (c) 13 (d) 12
The answer is 12 I guess

The forester plants the first sapling at the point where the pile is present. Hence he covers 0m while planting the first sapling. To plant the next sapling he has to cover 10m and comeback to the starting point covering another 10m. Hence 20m to plant sapling and come back to starting point. For next sapling he would take 40, the one after that 60 and so on.

Hence distance covered while planting n saplings ( and returning to starting point) is ->

0 + 20 + 40 + 60 + ......n terms = 1320

On solving using S.n formula we get n =12


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14-05-2005, 01:34 AM

Keep the work going junta.




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14-05-2005, 05:14 PM

so here we have to assume that the forester after planting the last sapling he wents back to his original position then only we will get the answer.

plz solve some more problems

What will be the maximum sum of 44, 42, 40 ...... ?

(a) 502 (b) 504 (c) 506 (d) none of these

The first and the third terms of an arithmetic progression are equal, respectively, to the first and the third term of the geometric progression, and the second term of the arithmetic progression exceeds the second term of the geometric progression by 0.25. Calculate the sum of the first five terms of the arithmetic progression if its first term is equal to 2.

(a) 2.25 or 25 (b) 2.5 or 27.5 (c) 1.5 (d) 3.25


Quote:
Originally Posted by Deathlord
The answer is 12 I guess

The forester plants the first sapling at the point where the pile is present. Hence he covers 0m while planting the first sapling. To plant the next sapling he has to cover 10m and comeback to the starting point covering another 10m. Hence 20m to plant sapling and come back to starting point. For next sapling he would take 40, the one after that 60 and so on.

Hence distance covered while planting n saplings ( and returning to starting point) is ->

0 + 20 + 40 + 60 + ......n terms = 1320

On solving using S.n formula we get n =12
   
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14-05-2005, 08:25 PM

Quote:
Originally Posted by catquery
so here we have to assume that the forester after planting the last sapling he wents back to his original position then only we will get the answer.

plz solve some more problems

What will be the maximum sum of 44, 42, 40 ...... ?

(a) 502 (b) 504 (c) 506 (d) none of these

The first and the third terms of an arithmetic progression are equal, respectively, to the first and the third term of the geometric progression, and the second term of the arithmetic progression exceeds the second term of the geometric progression by 0.25. Calculate the sum of the first five terms of the arithmetic progression if its first term is equal to 2.

(a) 2.25 or 25 (b) 2.5 or 27.5 (c) 1.5 (d) 3.25
Easy i guess...

1-ans is 506 the max sum is the sum of nos till 2, or IOW sum of first 22 even nos...
=n(n+1),=22*23=506

2-tried thru options, is the answer c-1.5..????
   
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