Loading....
Coupon Accepted Successfully!

 

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 ‘d’.
 
For example  Series (i) 1, 2, 3,…
Series (ii)  2, 5, 7, 10,…
Series (iii)  aa + da + 2d,…
 
Common Difference(d) of series (i) = 1
Common Difference(d) of series (ii) = 3
Common Difference(d) of series (iii) = d

nth term of an arithmetic progression
 
First term t1 = a = a + (1  1)d
Second term t2 = a + d = a + (2  1)d
Third term t3 = (a + d ) + d = a + 2d = a + (3  1)d
Fourth term t4 = a + 3d = a + (4  1)d

n
th term tn a + (n  1)d, where a is the first term, d is the common difference and n is the number of terms.

Important points
  • tn is also known as the general term of AP.
  • If in any question, some particular term is given (like t4 or t10), then we should assume those terms in the form of tn. However, if the total number of terms are given then we should assume the terms in the following way:
If three terms or any odd number of terms are involved, then we should assume these terms as a  daa + d and so on.

If four terms or any even number of terms are involved, then we should assume these terms as a – 3da  da + da + 3d 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  d, a , a + d.
Given is (a − d + a + a + d) = 27
So, a = 9
Also, (a − d)2 + a2 + (a + d)2 = 293
⇒ d2 = 25
⇒ d = ± 5
When d = + 5, then the terms are 4, 9, 14.
When d = − 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?
  1. 0
  2. –1
  3. 1
  4. Not unique
Solution
Given
t1 + t2 + … + t11 = t1 + t2 + … + t19 (for an AP)
Description: 2370.png 
⇒ 22a  + 110 d = 38a + 342d
⇒ 16a + 232 d = 0
⇒ 2a + 29 d = 0
Description: 2379.png 
⇒ S30 terms = 0
 

Properties of AP

If abcd,…are in AP, then
  1. a + k, b + k, c + k, d + k…will be in AP, where k is any constant.
  2. a  k, b  k, c  k, d  k…will be in AP, where k is any constant.
     
    In the above two cases, the common difference will be the same as earlier.
  3. ak, bk, ck, dk…will be in AP, where k is any constant.
     
    In this case, new common difference will be k times the earlier common difference.
  4. Description: 2388.png will be in AP, where k  0.
     
    In this case, new common difference will beDescription: 2397.png times the earlier common difference.
Example-3
If abc are in AP, then b + cc + aa + b will be in
  1. AP
  2. GP
  3. HP
  4. Cannot be determined uniquely
Solution
a, b, c are in AP, then a – (a + b + c), b – (a + b + c), c – (a + b + c) will be in AP.
⇒ −(b + c), − (a + c) and –(a + b) will be in AP.
⇒ (b + c), (a + c) and (a + b) will also be in AP.
Alternatively, let us assume abc to be 1, 2, 3. Then (b + c) = 5, (a + c) = 4 and (a + b) = 3, which are obviously in AP.
 
 
Example-4
If xyz are in GP, then 1/(1 + log10x), 1/(1 + log10y) and 1/(1 + log10z) will be in
  1. AP
  2. GP
  3. HP
  4. 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 + log10y) = 1/(1 + log10x) + 1/(1 + log10z)
Description: 2406.png 
= [2 + log10xz]/(1 + log10x) (1 + log10z)
Obviously this will not give us the answer.
 
Checking option (b),
[1/(1 + log10y)]2 = [1/(1 + log10x)] [1/(1 + log10z)]
= [1/(1 + log10(x z) + log10xz)]
Again, no solution is found.
 
Checking option (c),
1/(1+log10x ), 1/(1+log10y) and 1/(1+log10z) are in HP then 1+log10x, 1+log10y and 1 + log10z will be in AP.
So, log10x, log10y and log10z will also be in AP.
Hence, 2 log10y = log10x + log10z
⇒ y2 = 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 xy, z are in GP.
So, 1 + log10x = 1, 1 + log10y = 2 and 1 + log10z = 3.
⇒ Thus we find that since 1, 2 and 3 are in AP, we can assume that 1 + log10x , 1 + log10y and 1 + log10z are in AP.
Hence, by definition of an AP we have that 1/(1 + log10x), 1/(1 + log10y) and 1/(1 + log10z) are in AP. So, option (c) is the answer.
 

Sum of n terms of an arithmetic progression
Sn =Description: 2415.png[2a+(n1)d)], where n = number of terms, a = 1st term and d = common difference.
 
Example-5
Find the sum of AP 3, 5, 7,…50 terms and find its sum.
Solution
Here n = 50, d = 2 and a = 3.
Using formula Sn Description: 2424.png [2a + (n − 1)d)]
= 25 × [2 × 3 + (50 − 1)2)] = 25 × 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
Description: 2428.png
So, sum = average × number of numbers
= 52 × 50 = 2600
Hence, Sn = Description: 2437.png, 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 → 10, 17, 21,…94.
Here n = 13, d = 7 and a = 10
Using the formula for the sum
Sn = Description: 2446.png [2a + (n − 1)d)], sum = 676
Alternatively, using the average method, 
average = (1st number + last number)/2
Average = Description: 2450.png= 52
So, the sum = average × number of numbers
= 52 × 13 = 676
 

Example-7
Find the value of the expression
1 – 4 + 2 – 5 + 3 – 6 +…to 100 terms.
  1. −250
  2. −500
  3. −450
  4. –300
Solution
We can write the given expression (1 – 4 + 2 – 5 + 3 – 6 +…to 100 terms) as:
(1 + 2 + 3 +…to 50 terms) – (4 + 5 + 6 +…to 50 terms)
Both of these are AP’s with different values of ‘a’ and ‘d’.
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
=Description: 2459.png[2a + (n −1) d)], we get
Description: 2463.png [2.1 + (50 − 1).1] − Description: 2472.png [2.4 + (50 − 1).1]
= 25(2 + 49) – 25(8 + 49)
= 25(51 − 57) = −150
Alternatively, logically this question can be done a bit faster by assuming (1 − 4), (2 − 5), etc as one unit.
1 − 4 = 2 – 5 =…= −3.
Thus the above series is equivalent to a series of fifty (−3)’s added to each other.
So, (1 − 4) + (2 − 5) + (3 − 6) +…50 terms = − 3 × 50
= −150
 





Test Your Skills Now!
Take a Quiz now
Reviewer Name