# Arithmetic Progression

A sequence (finite or infinite) is called an **arithmetic progression**
(abbreviated A.P.) iff the difference of any term from its preceding term is
constant. This constant is usually denoted by d and is called **common
difference.**

### General term of an A.P.

T_{n}= a +(n -1) d.

Thus if a is the first term and d is the common difference of an A.P., then the A.P. is a, a +d, a +2 d, ..., a +(n -1) d or a, a +d, a +2 d, ... according as it is finite or infinite.

**Corollary.** If the last term of an A.P. consisting of n terms is denoted by l, then

l = a +(n -1) d.

**Note.** Three numbers a, b, c are in A.P. iff b -a = c -b i.e. iff a +c = 2b.

## Illustrative Examples

### Example

- A sequence {a
_{n}} is given by the formula a_{n}= 10 -3 n. Prove that it is an A.P. - Find the general term and 15th term of the sequence 3, 1, -1, -3, ....
- If the 9th term of an A.P. is 99 and 99th term is 9, find its 108th term.

### Solution

- Given a
_{n}= 10 -3 n => a_{n +1}= 10 -3 (n +1)

=> a_{n +1}-a_{n}= 10 -3 (n +1) -(10 -3 n) = -3, which is a constant. Therefore, the given sequence {a_{n}} is an A.P. - The given sequence is an A.P. with first term a = 3, and common difference d = -2.

Hence general term, T_{n}= a +(n -1) d = 3 +(n -1)(-2) = 5 -2n.

Putting n = 15, we get, T_{15}= 5 -2(15) = -25. - Let a be the first term and d be the common difference of given A.P.

Then T_{n}= a +(n -1) d.

Putting n = 9, 99 we get T_{9}= a +8d = 99, T_{99}= a +98d = 9

=> -90d = 90 => d = -1, a = 99 -8d = 107.

Hence T_{108}= a +107d = 107 +107(-1) = 0.

### Example

Show that the sequence {a_{n}} where a_{n} = 2n² +3 is not an A.P.

### Solution

Given a_{n} = 2n² +3 => a_{n+1} = 2(n +1)² +3.

= a_{n+1} -a_{n} = (2(n +1)² +3) -(2n² +3)

= (2(n² +2n
+1) +3) -(2n² +3) = 4n +2,

which depends upon n and is not constant.

Hence the given sequence is not an A.P.

### Example

In a sequence, the sum of first n terms is S_{n} = 2n² +3n +1 for
all values of n. Show that the series is an A.P. Find its 100th term.

### Solution

We have S_{n} = 2n² +3n +1,

S_{n+1} = 2(n -1)² +3(n -1) +1 = 2n² -n

T_{n} = S_{n} -S_{n-1} = (2n² +3n +1) -(2n² -n) = 4n +1

T_{n+1} -T_{n} = (4(n +1) +1) -(4n +1) = 4.

Hence the given series is an A.P. with common difference d = 4.

First term a = T_{1} = 4n +1 = 4(1) +1 = 5

Now 100th term = T_{100} = a +99d = 5 +99(4) = 401.

### Example

If a, b, c are in A.P., prove that the following are also in A.P.(b +c)² -a², (c +a)² -b², (a +b)² -c²

### Solution

(b +c)² -a², (c +a)² -b², (a +b)² -c² are in A.P.

if (b +c +a)(b +c -a), (c +a +b)(c +a -b)(a +b +c),

and (a +b -c) are in A.P.

i.e. if b +c -a, c +a -b, a +b -c are in A.P.

(dividing each term by a +b +c)

i.e. if (c +a -b) -(b +c -a) = (a +b -c) -(c +a -b)

i.e. if 2 (a -b) = 2 (b -c)

i.e. if a -b = b -c

i.e. if b -a = c -b

i.e. if a, b, c are in A.P. which is given to be true.

Hence (b +c)² -a², (c +a)² -b², (a +b)² -c² are in A.P.

## Exercise

- Find the

(i) n th term of the series 5 +1 -3 -7 ...

(ii) r th term of the series (a +d) +(a +3d) +(a +5d) +...

(iii) 100th term of the sequence 0, -3, -6, -9, ...

(iv) m th term of the sequence n -1, n -2, n -3, ...

(v) Eighteenth term of the sequence 9, 5, 1, ... - Find first four terms of the series

(i) (6n +1) (ii) whose nth term is 3 -2n

(iii) given that series is A.P. with a = 14/2, d = 3/2. - If a, b are positive real numbers, show that the sequence

log a, log (ab), log (ab²), log (ab³), ... is an A.P.

Also find its general term. - Which term of the series

(i) 2 +5 +8 +11 +... is 53

(ii) 20 +18 +16 +... is -2

(iii) 4 +0 -4 -8 ... is -392? - Which term of the sequence 20, 19 is the first negative term?
- Which term of the sequence 9 -8i, 8 -6i, 7 -4i, ... is

(i) real (ii) purely imaginary? - (i) Find the arithmetic progression of 6 terms if first term is 2/3 and the last is 22/3.

(ii) Find the number of terms in the series 5 +8 +11 +14 +... if the last term is 95. - (i) Determine the 25th term of an A.P. whose 9th term is -6 and common difference is 5/4.

(ii) The third term of an A.P. is 25 and tenth term is -3. Find the first term and the common difference. - (i) If 9th term of an A.P. is zero, prove that 29th term is double the 19th term.

(ii) If 7 times the 7th term of an A.P. is equal to 11 times its 11th term show that 18th term of this A.P. is zero. - Second, 31st and last terms of an A.P. are 31/4, 1/2, -13/2 respectively. Find the number of terms in the A.P.
- Determine k so that k +2, 4k -6 and 3k -2 are three consecutive terms of an A.P.
- (i) If the nth term of a sequence is an expression of first degree in n,
show that it is an A.P.

[**Hint.**Take T_{n}= an +b]

(ii) A sequence {a_{n}} is given by a_{n}= n² -1, nN. Show that it is not an A.P. - A person was drawing a monthly salary of Rs 2500 in 11th year of service and a salary of Rs 2900 in the 19th year. Given that pension is half the salary at retirement time, find his monthly pension if he had put in 25 years of service before retirement. Assume that annual increment is constant.
- A person purchases a VCR for Rs 16000. Its life is estimated to be 20 years. If the yearly depreciation is assumed to be constant, find the rate of depreciation and the price after 8 years.
- If a, b, c are in A.P. show that

(i) (a -c)² = 4(b² -a c) (ii) (a -c)² = 4(a -b)(b -c) - For all real a, b, prove that (a -b)², a² +b² and (a +b)² are in A.P.
- How many numbers of two digits are divisible by 6?

[**Hint:**The first two digit number divisible by 6 is 12 and the last two digit number divisible by 6 is 96.]

## Answers

**1.**(i) 9 -4n (ii) a +(2r -1)d (iii) -297

(iv) n -m (v) -59

**2.**(i) 7, 13, 19, 25 (ii) 1, -1, -3, -5 (iii) 7, 12

**3.**log (a b

^{n-1})

**4.**(i) 18 th (ii) 12 th (iii) 100 th

**5.**28th

**6.**(i) 5th (ii) 10th

**7.**(i)2/3, 2, 10/3, 14/3, 6, 22/3 (ii) 31

**8.**(i) 14 (ii) 33, -4

**10.**59

**11.**3

**13.**Rs 1600

**14.**Rs 800, Rs 9600

**17.**15