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Mathematics: Post your doubts here!

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okay ppl listen up :p

do Q8 from this one -__-
i'm dead exhausted and can't solve it so help :O
 

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okay ppl listen up :p

do Q8 from this one -__-
i'm dead exhausted and can't solve it so help :O
ok here goes nothing
1 on first day the amount donated is 5/100 *1000= 50
on second day the amount donated is 5/100 *2000=100 hence we can see that d=50 a=50 now use formulae S40=40/2(100 + (39)50) =40000
 

asd

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okay ppl listen up :p

do Q8 from this one -__-
i'm dead exhausted and can't solve it so help :O
I solved it just yesterday.
Now write down the numbers for the first model: they are going to be 1000,2000,3000,...
5 percent donated everyday, 50,100,150.
first term is 50, common difference is 50.
solve for the sum of 40 terms. you get 41000
For the second model, do the same. 1000,1100,1210,1331...
r= 1.1
S40=1000(r^40-1)/(r-1)
what ever you get just take the 5 percent of that.
 
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I solved it just yesterday.
Now write down the numbers for the first model: they are going to be 1000,2000,3000,...
5 percent donated everyday, 50,100,150.
first term is 50, common difference is 50.
solve for the sum of 40 terms. you get 41000
For the second model, do the same. 1000,1100,1210,1331...
r= 1.1
S40=1000(r^40-1)/(r-1)
what ever you get just take the 5 percent of that.
got it was using a instead of 2a :p
 
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Q. A geometric progression has the first term 4, and common ration (2x-3) , find the set of values of x for which the geometric progression has a sum to infinity.

How o_O?
 
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Q. A geometric progression has the first term 4, and common ration (2x-3) , find the set of values of x for which the geometric progression has a sum to infinity.

How o_O?
r<1 if sum to infinity exists
solve this inequality using the vlaue of r u have : - (2x-3) and u'll get ur range... i think :p
 
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