# Arithmetic Progression

A succession of numbers is said to be in arithmetic progression (AP) if the difference between any term and the previous term is constant throughout. In other words, the difference between any of the two consecutive terms should be the same. This difference which is common between any two consecutive terms is known as common difference of this AP and is denoted by*â€˜dâ€™.*

For example â†’ Series (i) 1, 2, 3,â€¦

Series (ii) â†’ 2, 5, 7, 10,â€¦

Series (iii) â†’

*a*,*a*+*d*,*a*+ 2*d*,â€¦Common Difference(

*d*) of series (i) = 1Common Difference(

*d*) of series (ii) = 3Common Difference(

*d*) of series (iii) =*d*

nth term of an arithmetic progression

nth term of an arithmetic progression

First term

*t*_{1}=*a*=*a*+ (1 âˆ’ 1)*d*Second term

*t*_{2}=*a*+*d*=*a*+ (2 âˆ’ 1)*d*Third term

*t*_{3}= (*a*+*d*) +*d*=*a*+ 2*d*=*a*+ (3 âˆ’ 1)*d*Fourth term

*t*_{4}=*a*+ 3*d*=*a*+ (4 âˆ’ 1)*d**th term*

n

n

*t*

*=*

_{n }*a*+ (

*n*âˆ’ 1)

*d,*where

*a*is the first term,

*d*is the common difference and

*n*is the number of terms.

**Important points**

*t*_{n}_{ }is also known as the general term of AP.- If in any question, some particular term is given (like
*t*_{4}or*t*_{10}), then we should assume those terms in the form of*t*. However, if the total number of terms are given then we should assume the terms in the following way:_{n}

If three terms or any odd number of terms are involved, then we should assume these terms as

If four terms or any even number of terms are involved, then we should assume these terms as

*a*âˆ’*d*,*a*,*a*+*d*and so on.If four terms or any even number of terms are involved, then we should assume these terms as

*a*â€“ 3*d*,*a*âˆ’*d*,*a*+*d*,*a*+ 3*d*and so on.Example-1

The sum of three numbers in an AP is 27 and the sum of their squares is 293. Find the numbers.

Solution

Let the numbers be

*a*âˆ’*d, a , a*+*d*.Given is (

*a*âˆ’*d*+*a*+*a*+*d*) = 27So,

*a*= 9Also, (

*a*âˆ’*d*)^{2}+*a*^{2}+ (*a*+*d*)^{2}= 293â‡’

*d*^{2}= 25â‡’

*d*= Â± 5When

*d*= + 5, then the terms are 4, 9, 14.When

*d*= âˆ’ 5, then the terms are 14, 9, 4.Alternatively, this question can be worked out very easily with the help of options.

Example-2

If the sum of the first 11 terms of an arithmetic progression equals that of the first 19 terms, then what is the sum of the first 30 terms?

- 0
- â€“1
- 1
- Not unique

Solution

Given

*t*

_{1}+

*t*

_{2}+ â€¦ +

*t*

_{11}=

*t*

_{1}+

*t*

_{2}+ â€¦ +

*t*

_{19}(for an AP)

â‡’ 22

*a*+ 110*d*= 38*a*+ 342*d*â‡’ 16

*a*+ 232*d*= 0â‡’ 2

*a*+ 29*d*= 0â‡’ S

_{30 terms}= 0

# Properties of AP

If

*a*,*b*,*c*,*d*,â€¦are in AP, then*a*+ k,*b*+ k,*c*+ k,*d*+*k*is any constant.*a*âˆ’*b*âˆ’ k,*c*âˆ’ k,*d*âˆ’ kâ€¦will be in AP, where k is any constant.*a*k,*b*k,*c*k,*d*kâ€¦will be in AP, where k is any constant.*k*times the earlier common difference.- will be in AP, where k â‰ 0.

Example-3

If

*a*,*b*,*c*are in AP, then*b*+*c*,*c*+*a*,*a*+*b*will be in- AP
- GP
- HP
- Cannot be determined uniquely

Solution

*a, b, c*are in AP, then

*a*â€“ (

*a*+

*b*+

*c*),

*b*â€“ (

*a*+

*b*+

*c*),

*c*â€“ (

*a*+

*b*+

*c*) will be in AP.

â‡’ âˆ’(

*b*+*c*), âˆ’ (*a*+*c*) and â€“(*a*+*b*) will be in AP.â‡’ (

*b*+*c*), (*a*+*c*) and (*a*+*b*) will also be in AP.Alternatively, let us assume

*a*,*b*,*c*to be 1, 2, 3. Then (*b*+*c*) = 5, (*a*+*c*) = 4 and (*a*+*b*) = 3, which are obviously in AP.Example-4

If

*x*,*y*,*z*are in GP, then 1/(1 + log_{10}*x*), 1/(1 + log_{10}*y*) and 1/(1 + log_{10}*z*) will be in- AP
- GP
- HP
- Cannot be determined uniquely

Solution

Let us go through the options

Checking option (a), the three will be in AP if the 2nd expression is the average of the 1st and the 3rd expressions. This can be mathematically written as

2/(1 + log

_{10}*y*) = 1/(1 + log_{10}*x*) + 1/(1 + log_{10}*z*)= [2 + log

_{10}*xz*]/(1 + log_{10}*x*) (1 + log_{10}*z*)Obviously this will not give us the answer.

Checking option (b),

[1/(1 + log

_{10}*y*)]^{2}= [1/(1 + log_{10}*x*)] [1/(1 + log_{10}*z*)]= [1/(1 + log

_{10}(*x*+*z*) + log_{10}*xz*)]Again, no solution is found.

Checking option (c),

1/(1+log

_{10}*x*), 1/(1+log_{10}*y*) and 1/(1+log_{10}*z*) are in HP then 1+log_{10}*x*, 1+log_{10}*y*and 1 + log_{10}*z*will be in AP.So, log

_{10}*x*, log_{10}*y*and log_{10}*z*will also be in AP.Hence, 2 log

_{10}*y*= log_{10}*x*+ log_{10}*z*â‡’

*y*^{2}=*xz*which is given.So, (c) is the answer.

Alternatively, we can also apply the following process:

Assume

*x*= 1,*y*= 10 and*z*= 100 as*x*,*y*,*z*are in GP.So, 1 + log

_{10}*x*= 1, 1 + log_{10}*y*= 2 and 1 + log_{10}*z*= 3.â‡’ Thus we find that since 1, 2 and 3 are in AP, we can assume that 1 + log

_{10}*x*, 1 + log_{10}*y*and 1 + log_{10}*z*are in AP.Hence, by definition of an AP we have that 1/(1 + log

_{10}*x*), 1/(1 + log_{10}*y*) and 1/(1 + log_{10}*z*) are in AP. So, option (c) is the answer.*Sum of n terms of an arithmetic progression*

S

_{n }=[2*a*+(*n*âˆ’1)*d*)], where*n*= number of terms,*a*= 1st term and*d*= common difference.Example-5

Find the sum of AP 3, 5, 7,â€¦50 terms and find its sum.

Solution

Here

*n*= 50,*d*= 2 and*a*= 3.Using formula S

_{n }= [2*a*+ (*n*âˆ’ 1)*d*)]= 25 Ã— [2 Ã— 3 + (50 âˆ’ 1)2)] = 25 Ã— 104 = 2600

However, we can find out the sum of any AP in a better way through average also.

The last term of this series = 101, so the average

=

So, sum = average Ã— number of numbers

= 52 Ã— 50 = 2600

Hence, S

_{n}= , where F is the first term, L is the last term and*n*is the number of terms of that AP.Example-6

What is the sum of all the two-digit numbers which when divided by 7 gives a remainder of 3? (CAT 2003)

Solution

This series is like â†’ 10, 17, 21,â€¦94.

Here

*n*= 13,*d*= 7 and*a*= 10Using the formula for the sum

S

_{n}= [2*a*+ (*n*âˆ’ 1)*d*)], sum = 676Alternatively, using the average method,

average = (1st number + last number)/2

Average = = 52

So, the sum = average Ã— number of numbers

= 52 Ã— 13 = 676

Example-7

Find the value of the expression

1 â€“ 4 + 2 â€“ 5 + 3 â€“ 6 +â€¦to 100 terms.

1 â€“ 4 + 2 â€“ 5 + 3 â€“ 6 +â€¦to 100 terms.

- âˆ’250
- âˆ’500
- âˆ’450
- â€“300

Solution

We can write the given expression (1 â€“ 4 + 2 â€“ 5 + 3 â€“ 6 +â€¦to 100 terms) as:

(1 + 2 + 3 +â€¦to 50 terms) â€“ (4 + 5 + 6 +â€¦to 50 terms)

Both of these are APâ€™s with different values of â€˜

*a*â€™ and â€˜*d*â€™.*a*= 1,

*n*= 50 and

*d*= 1 for 1st series and

*a*= 4,

*n*= 50 and

*d*= 1 respectively.

Using the formula for the sum of AP

=[2

*a*+ (*n*âˆ’1)*d*)], we get [2.1 + (50 âˆ’ 1).1] âˆ’ [2.4 + (50 âˆ’ 1).1]

= 25(2 + 49) â€“ 25(8 + 49)

= 25(51 âˆ’ 57) = âˆ’150

Alternatively, logically this question can be done a bit faster by assuming (1 âˆ’ 4), (2 âˆ’ 5), etc as one unit.

1 âˆ’ 4 = 2 â€“ 5 =â€¦= âˆ’3.

Thus the above series is equivalent to a series of fifty (âˆ’3)â€™s added to each other.

So, (1 âˆ’ 4) + (2 âˆ’ 5) + (3 âˆ’ 6) +â€¦50 terms = âˆ’ 3 Ã— 50

= âˆ’150