Progressions - Model Problems - 6

Complex problems in progressions

Prob 1: Find the sum of the series

S = 1²-2²+3²-4²+…-2002²+2003²

We need transform the given series into

S = (1-2)(1+2) + (3-4)(3+4)+…+(2001-2002)(2001+2002)+2003²
=> (-1)(3) + (-1)(7) + (-1)(11) +(-1)(15)+…+(-1)(4003) + 2003²

=> (-1)[3+7+11+…+4003]+2003²
So we have an arithmetic progression in the brackets with a =3, d = 4 and l = 4003 and n = 1001 terms.

Sum to n terms in AP is
Sn = ½ n(a + l)

Hence the sum of 1001 terms in the brackets = 1001*[3+4003]/2
= 1001*4006/2 = 1001*2003

So S = -1001*2003 + 2003²
= 2003(2003-1001) = 2003*1002 = 2007006


Prob 2: Find the value of n for which
704 + ½ *(704)+1/4 * (704) +… up to n terms =
1984 – ½ *(1984) + ¼* (1984)- … up to n terms

Actually there is no complexity in the problem. But for a first look it looks to be a complex problem. You have to identify both LHS and RHS as geometric progressions.
For LHS a = 704 and r = ½.
For RHS a = 1984 and r = -1/2

sum of n terms of GP
Sn = a(1-rⁿ)/(1-r)

704[1 – (1/2)ⁿ]/(1 – ½ ) = 1984 [1 – (-1/2)ⁿ]/(1 – (-½ ))
704*2(1 – 1/2ⁿ] = 1984*2/3*[1 – (-1) ⁿ/2ⁿ]
704*6(1 – 1/2ⁿ] = 1984*2[1 – (-1) ⁿ/2ⁿ]
4224 – 4224/2ⁿ = 3968 - 3968(-1) ⁿ/2ⁿ
4224-3968 = 4224/2ⁿ - 3968(-1) ⁿ/2ⁿ
256 = 4224/2ⁿ - 3968(-1) ⁿ/2ⁿ
128 = 2112/2ⁿ - 1984(-1) ⁿ/2ⁿ

If n is assumed as odd
128 = (2112+1984)/ 2ⁿ
=> 2ⁿ = 4096/128 = 1024/32 = 128/4 = 32
=> 2ⁿ = 32
n = 5

If n is assumed as even

128 = (2112 – 1984)/ 2ⁿ = 128/2ⁿ
=> 2ⁿ = 128/128 = 1
Implies n = 0

Hence the answer is n =5 or n = 0. The logical answer is 5 as n = 0 is a trivial solution..