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How many terms at the maximum, of the progression 2,5,8,........can be

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How many terms at the maximum, of the progression 2,5,8,........can be [#permalink] New post   11 May 2020, 23:34
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How many terms at the maximum, of the progression 2,5,8,........can be considered, if the sum of the terms is to be less than 3000?
A.60
B.44
C.68
D.70
E.67

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Re: How many terms at the maximum, of the progression 2,5,8,........can be [#permalink] New post   11 May 2020, 23:55
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MBADream786 wrote:
How many terms at the maximum, of the progression 2,5,8,........can be considered, if the sum of the terms is to be less than 3000?
A.60
B.44
C.68
D.70
E.67


CONCEPT:Sum of an Arithmetic Progression = (n/2)*(First-Term + Last-Term) = (n/2)*[2a+(n-1)*d]


Here, First term, a = 2
Common Difference, d = 5-2 = 8-5 = 3

i.e. Sum of an Arithmetic Progression = (n/2)*[2*2+(n-1)*3] < 3000 (Given)

i.e. Sum of an Arithmetic Progression = (n/2)*[4+(3n-3)] < 3000

i.e. Sum of an Arithmetic Progression = (n/2)*(3n+1) < 3000

i.e. (n)*(3n+1) < 6000

Let,s Check Options:

A.60, i.e. n = 60, (n)*(3n+1) = 60*181 > 6000 OUT and all options other than B are OUT too
B.44, i.e. n = 4, (n)*(3n+1) = 44*133 = 5852 < 6000 CORRECT
C.68
D.70
E.67

ANswer: Option B
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Re: How many terms at the maximum, of the progression 2,5,8,........can be [#permalink] New post   12 May 2020, 00:05
Thanks a lot for the quick reply.
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Re: How many terms at the maximum, of the progression 2,5,8,........can be [#permalink] New post   12 May 2020, 00:18
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Tn= 2+(n-1)*3
=3n-1

S= 2+5+8+...+(3n-1)
=n/2[2*2+(n-1)*3]
=n/2[3n+1]

now,n/2[3n+1] <3000
or,n(3n+1)<6000

putting,n=60 , 60*181= 10860 (rejected)
n=44, 44*133 = 5852 ( <6000)
we don't need to check others because rest options are far away from 44.

correct answer B

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Re: How many terms at the maximum, of the progression 2,5,8,........can be [#permalink]
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