IIT JEE , AIEEE Forum - Mathematics , Algebra

varshavallig

Let r V denote the sum of the first r terms of an arithmetic progression (A.P.) whose first term is r and the

common difference is (2r-1) .

Let Tr= V_r+1 -V_r -2 andQ_r= T_r+1- T_r for r = 1, 2,?.

find V_r!!!!!!!!!!!!!!!!!!

plz reply soon

rates assured

!!!!!!!!!!!!!!!!!!

Avinash_Bhat

YOUR QUESTION IS JUST TO FIND Vr

First term = r

Common Difference = ( 2r - 1 )

Sum of first r terms = Vr

= ( r / 2 )( 2r + (r - 1)(2r - 1) )

cuteprincess

hey what is Q there????!!!

varshavallig

but thats not the right ans .
me too did the same way.in this method v get a cube term.but in the solution there is no such term.

the ans is Vr= r^2-r^2/2 +r/2
hows that possible.
!!!!!!!!!!!!!!!!!!!!!!!

priyesh

@varshavallig

then ans given in your book is wrong!!!!!!!!!!!!!

because

putting r = 2 in ur ans sum comes out as 3

but sum of first two terms is 2 + 5 = 7

sugeet

r² + r/2(r-1)(2r-1)

got it by sum of r terms formulae

hope not wrong

varshavallig

it was given here: http://www.careerlauncher.com/common/analyses/2007_IITJEE_paper1.pdf
plz have a look.
!!!!!!!!!!!!!!!!!!

sunip_the_mini_mastermind

u wrote out the wrong question ...the question is to find_{[ ]}

^{[ ]} V_r

aritra

its all very simple, folks

the r

2 term will be r3

varshavallig

@ sunip_the_mini_mastermind
but there in the step for finding Vr ,it is diff from what v got ,so if the 1st step is wrong,obviously sum of Vr is also wrong.
!!!!!!!!

priyesh

Vr = r/2(2r + (r-1)(2r-1)] = 1/2[2r^3 - r^2 + r)

Vr = 1/2[ 2

r^3 -

r^2 +

r]

= 1/2[2(n(n+1)/2)^2 - n(n+1)(2n+1)/6 + n(n+1)/2

= 1/12[n(n+1)(3n^2 + n + 2)]

priyesh

@sugeet ur signature width is very large please reduce it

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