Progressions - Model Problems - 1

Chapter: Sequences, Series and Progressions

Arithmetic Progressions

Finding nth term of arithmetic progessions
nth term = a +(n-1)d

Prob: Find the 20th term of the A.P. starting with 4 and common difference 5.

a = 4
d = 5
n = 20

20th term = 4 + (20-1)*5 = 4+19*5 = 4 + 95 = 99.

Finding sum to n terms of AP

Sn = ½ n(a + l)
Sn = ½ n{2a+(n-1)d}


Prob: Find the sum to 20th term of the A.P. starting with 4 and common difference 5.

We found in the earlier problem the 20th term.

a = 4
n = 20
l = 99

Sn = ½ *20(4+99) = 10*103 = 1030.

Insertion of arithmetic means

Prob: Insert 4 arithmetic means between 4 and 19.

We have a = 4, l = 19 and 4 arithmetic means are to be inserted. It means the last term l is the 6th term.

19 = 4+5d
d = 3
the firm A.M. = second term = 4+3 = 7
second A.M. = 7+3 = 10
third A.M. = 10+3 = 13
fourth A.M. = 13+3 = 16

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..

Progressions - Model Problems - 7

1. The third term of a geometric progression is 4. The product of the first five terms is:

a. 4³
b. 4⁴
c. 4⁵
d. none of these

(JEE 1982)

Answer ©

Select the five terms as a/r², a/r,a, ar,ar².

Product of the five terms is a/r² * a/r * a *ar*ar² = a⁵

As the third term is 4, a = 4
Hence product is 4⁵

2. The sum of integers from 1 to 100 that are divisible by 2 or 5 is ---------------------.
(JEE 1984)

Answer: 3050

Integers divisible by 2 are 2,4,6,…,100 ….(50 numbers)
Integers divisible by 5 are 5, 10, 15,…100 (20 numbers)

Integers divisible by 10 will be in integers divisible by 2. We can remove them from integers divisible by 5 to find out the sum required

So integers for which sum is to be taken 2,4,6…,100 and
5,15,25,…,95

First progression sum 2(sum of 1,2,3,…,50)
= 2*50*51/2 = 2550

Second progression is an A.P. with d = 10 and a = 5

Hence sum = n/2(a+l) = 10/2(100) = 500

Total sum = 2550 +500 = 3050