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by shashank.ism » Tue Feb 09, 2010 7:19 am
If the sum of the first 11 terms of an arithmetic progression equals that of the first l9 terms, then what is the sum of the first 30 terms?

a 0
b -1
c 1
d 2
e Not unique
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by ajith » Tue Feb 09, 2010 11:37 am
shashank.ism wrote:If the sum of the first 11 terms of an arithmetic progression equals that of the first l9 terms, then what is the sum of the first 30 terms?

a 0
b -1
c 1
d 2
e Not unique
11/2 (2a+10d) = 19/2 (2a +18d)

11a+ 55d = 19a+ 171d

8a +116d =0
2a+29d =0


Sum of first 30 terms = 15*(2a+29d) =0
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by harsh.champ » Tue Feb 09, 2010 11:52 am
shashank.ism wrote:If the sum of the first 11 terms of an arithmetic progression equals that of the first l9 terms, then what is the sum of the first 30 terms?

a 0
b -1
c 1
d 2
e Not unique
Sum of 1st 11 terms = 11a + n x 10(10 + 1)/2
Sum oif 1st 19 terms = 19a + n x 18(18 + 1)/2

Since they are equal,hence from 1 & 2 we get that 8a = -116d
Hence, we have to find 30a + n x 29(29 + 1)/2 ,put 8a = -116d we get the sum as o.ans A
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by komal » Wed Feb 17, 2010 11:38 am
shashank.ism wrote:If the sum of the first 11 terms of an arithmetic progression equals that of the first l9 terms, then what is the sum of the first 30 terms?

a 0
b -1
c 1
d 2
e Not unique
Sum of the first 11 terms of an A.P. for which first term is a and common
difference is d will be 11/2(2a+10d) and for 19 terms it will be 19/2(2a+18d).
According to problem 11/2(2a+10d)=19/2(2a+18d)
22a+110d=38a+342d
16a=-232d --> 2a=-29d
Now we have to find out sum of first 30 terms that will be
15(2a+29d)=30a+435d --> 15*(-29)d+435d = 0. Hence answer option is 1.
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