The question is about Sum of a progression. A term is given and we need to find out the order of the term in the series. With some simple but very powerful ideas, one can cut down on a lot of working when it comes to progressions. For example, anchoring a progression around its middle term can be very useful. CAT Exam tests the idea of progression often in the CAT Quantitative Aptitude section and this could also be tested in DI LR section of the CAT Exam as a part of a puzzle.
Question 12: If Sn = n3 + n2 + n + 1 , where Sn denotes the sum of the first n terms of a series and tm = 291, then m is equal to?
Sn - Sn = tn
Substitute m instead of n
Sm - Sm-1 = tm
We know that Sn = n3 + n2 + n + 1
Hence m3 + m2 + m + 1
m3 + m2 + m + 1 - [(m-1)3 + (m-1)2 + (m-1) + 1 ] = 291
m3 + m2 + m + 1 - [m3 - (3m)2 + 3m - 1 + m2 - 2m + 1 + m - 1 + 1] = 291
1 + (3m)2 + 3m + 1 -2m - 1 = 291
(-3m)2 + m - 290 = 0
(3m)2 - m + 290 = 0
Solving above equation we get m = -29, 30
M cannot be negative
Hence m = 30
The question is "If Sn = n3 + n2 + n + 1 , where Sn denotes the sum of the first n terms of a series and tm = 291, then m is equal to?"
Choice B is the correct answer.
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