AP -I
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maihuna
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There are two sets of numbers each consisting of 3 terms and sum of each set is 15. The common difference of the first is greater by 1 than the common difference of the second, and the product of first set is to the product of the second set as 7/8, what are the numbers.
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tohellandback
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You have already posted your 2nd question in some other thread.
Answer to your 1st question:
sum of odd terms=24
sum of even terms=30
let there be n terms
sum of odd terms:
the first term is a, last term is "last term of the series-common difference" i.e a+10.5-d
n/4(a+a+10.5-d)=24
n(2a+10.5-d)=96------------1
sum of even terms:
1st term is a+d, last term is a+10.5
n/4(a+d+a+10.5)=30
n(2a+10.5+d)=120----------2
subtract 1 from 2
2nd=24
nd=12
d=12/n--------------3
also given that
a+(n-1)d= a+10.5, where d is the common difference
d=10.5/(n-1)
d=10.5/n-1=12/n
10.5n=12n-12
1.5n=12
n=8
Answer B
Answer to your 1st question:
sum of odd terms=24
sum of even terms=30
let there be n terms
sum of odd terms:
the first term is a, last term is "last term of the series-common difference" i.e a+10.5-d
n/4(a+a+10.5-d)=24
n(2a+10.5-d)=96------------1
sum of even terms:
1st term is a+d, last term is a+10.5
n/4(a+d+a+10.5)=30
n(2a+10.5+d)=120----------2
subtract 1 from 2
2nd=24
nd=12
d=12/n--------------3
also given that
a+(n-1)d= a+10.5, where d is the common difference
d=10.5/(n-1)
d=10.5/n-1=12/n
10.5n=12n-12
1.5n=12
n=8
Answer B
The powers of two are bloody impolite!!












