5.3 THE Nth TERM OF AN ARITHMETIC PROGRESSION:
If an is the nth term, a1 is the first term, n is the number of terms in the sequence and d is a common difference then the nth term of an Arithmetic Progression will be
Example: Find the 11th term of the AP: 24, 20, 16,…
Solution: Given a = 24, n = 11, d = 20 – 24 = – 4
an = a + (n – 1)d
a11 = 24 + (11−1) – 4 ⇒ 24 + (10) – 4
= 24 – 40 ⇒ – 16
EXERCISE 5.1
1.In which of the following situations, does the list of numbers involved make an Arithmetic Progression, and why.
( i) The taxi fare after each km when the fare is Rs 15 for the first km and Rs 8 for each extra km.
Solution: Taxi fare for 1st km = Rs 15, Taxi fare after 2 km = 15 + 8 = Rs 23
Taxi fare after 3 km = 23 + 8 = Rs 31
Taxi fare after 4 km = 31 + 8 = Rs 39
Therefore, the sequence is 15, 23, 31, 39…
It is an arithmetic progression because difference between any two consecutive terms is equal which is 8. ( 23 – 15 = 8, 31 – 23 = 8, 39 – 31 = 8, … )
( ii)The amount of air present in a cylinder when a vacuum pump removes 1 / 4th of the air remaining in the cylinder at a time.
Solution: Let the volume of air in a cylinder, initially, be V litres.
In each stroke, the vacuum pump removes 1 / 4th of air remaining in the cylinder at a time. Or we can say, after every stroke, 1 − 1 / 4 = 3 / 4th part of air will remain.
Therefore, volumes will be V, (3V / 4) , ( 3V / 4)2 , ( 3V / 4)3… and so on
Clearly, we can see here, the adjacent terms of this series do not have the common difference between them. Therefore, this series is not an A.P.
( iii) The cost of digging a well after every meter of digging, costs Rs 150 for the first meter and rises by Rs 50 for each subsequent meter.
Solution:Cost of digging 1 meter of well = Rs 150
Cost of digging 2 meters of well = 150 + 50 = Rs 200
Cost of digging 3 meters of well = 200 + 50 = Rs 250
Therefore, we get a sequence of the form 150, 200, 250…
It is an Arithmetic Progression because difference between any two consecutive terms is equal. ( 200 – 150 = 50) , (250 – 200 = 50 ) and so on
Here, difference between any two consecutive terms which is also called common difference is equal to 50.
( iv) The amount of money in the account every year, when Rs 10,000 is deposited at compound Interest at 8% per annum.
Solution:We know that if Rs. P is deposited at r% compound interest per annum for n years, the amount of money will be: P ( 1 + r / 100) n
Therefore, after each year, the amount of money will be;
10000 ( 1 + 8 / 100), 10000 ( 1 + 8 / 100) 2, 10000 ( 1 + 8 / 100) 3 ……
Clearly, the terms of this series do not have the common difference between them. Therefore, this is not an A.P.
- Write first four terms of the AP, when the first term a and common difference d are given as:
( i) a = 10, d = 10
Solution: First term = a = 10, d = 10
Second term = a + d = 10 + 10 = 20
Third term = second term + d = 20 + 10 = 30
Fourth term = third term + d = 30 + 10 = 40
Therefore, first four terms are: 10, 20, 30, 40
( ii) a = − 2, d = 0.
Solution: First term = a = –2 , d = 0.
Second term = a + d = –2 + 0 = –2.
Third term = second term + d = –2 + 0 = –2.
Fourth term = third term + d = –2 + 0 = –2.
Therefore, first four terms are: –2, –2, –2, –2.
( iii) a = 4, d = − 3
Solution:First term = a = 4, d = –3
Second term = a + d = 4 – 3 = 1
Third term = second term + d = 1 – 3 = –2
Fourth term = third term + d = –2 – 3 = –5
Therefore, first four terms are: 4, 1, –2, –5
( iv) a = − 1, d = ½
Solution: Let us consider, the Arithmetic Progression series be a1, a2, a3, a4, a5 …
a2 = a1 + d = − 1 + 1 / 2 = − 1 / 2
a3 = a2 + d = − 1 / 2 + 1 / 2 = 0
a4 = a3 + d = 0 + 1 / 2 = 1 / 2
Thus, the A.P. series will be − 1, − 1 / 2, 0, 1 / 2
And First four terms of this A.P. will be − 1, − 1 / 2, 0 and 1 / 2.
( v) a = − 1.25, d = − 0.25.
Solution: First term = a = –1.25, d = –0.25.
Second term = a + d = –1.25 – 0.25 = –1.50.
Third term = second term + d = –1.50 – 0.25 = –1.75.
Fourth term = third term + d = –1.75 – 0.25 = –2.00.
Therefore, first four terms are: –1.25, –1.50, –1.75, –2.00.
- For the following AP’s, write the first term and the common difference.
( i) 3, 1, –1, –3 …
Solution: First term = a = 3,
Common difference ( d) = Second term – first term = Third term – second term and so on
Therefore, Common difference ( d) = 1 – 3 = –2
( ii) –5, –1, 3, 7…
Solution: First term = a = –5
Common difference ( d) = Second term – First term
= Third term – Second term and so on
Therefore, Common difference ( d) = –1 – ( –5) = –1 + 5 = 4
( iii) 1 / 3, 5 / 3, 9 / 3, 13 / 3 ….
Solution: First term, a = 1 / 3
Common difference, d = Second term – First term
⇒ 5 / 3 – 1 / 3 = 4 / 3
( iv) 0.6, 1.7, 2.8, 3.9 …
Solution: First term = a = 0.6
Common difference ( d) = Second term – First term
= Third term – Second term and so on
Therefore, Common difference ( d) = 1.7 − 0.6 = 1.1
- Which of the following are AP’s . If they form an AP, find the common difference d and write three more terms.
( i) 2, 4, 8, 16…
Solution: Here, the common difference is;
a2 – a1 = 4 – 2 = 2
a3 – a2 = 8 – 4 = 4
a4 – a3 = 16 – 8 = 8
Since, an + 1 – an or the common difference is not the same every time.
Therefore, the given series are not forming an A.P.
( ii) 2, 5 / 2, 3, 7 / 2 ….
Solution: Here, the common difference is;, a2 – a1 = 5 / 2 − 2 = 1 / 2
a3 – a2 = 3 − 5 / 2 = 1 / 2
a4 – a3 = 7 / 2 − 3 = 1 / 2
Since, an + 1 – an or the common difference is same every time.
Therefore, d = 1 / 2 and the given series are in A.P.
The next three terms are;
a5 = 7 / 2 + 1 / 2 = 4
a6 = 4 + 1 / 2 = 9 / 2
a7 = 9 / 2 + 1 / 2 = 5
( iii) − 1.2, − 3.2, − 5.2, − 7.2…
Solution: Here, the common difference is;
− 3.2 − ( − 1.2) = − 2
− 5.2 − ( − 3.2) = − 2
− 7.2 − ( − 5.2) = − 2
Common difference ( d) = − 2
Since, an + 1 – an or the common difference is same every time.
Fifth term = − 7.2 – 2 = − 9.2 Sixth term = − 9.2 – 2 = − 11.2
Seventh term = − 11.2 – 2 = − 13.2
Therefore, next three terms are − 9.2, − 11.2 and − 13.2
( iv) − 10, − 6, − 2, 2…
Solution: Here, the common difference is;
− 6 − ( − 10) = 4 − 2 − ( − 6) = 4
2 − ( − 2) = 4
Common difference ( d) = 4
Since, an + 1 – an or the common difference is same every time.
Fifth term = 2 + 4 = 6 Sixth term = 6 + 4 = 10
Seventh term = 10 + 4 = 14
Therefore, next three terms are 6, 10 and 14.
( v) 3, 3 + √ 2, 3 + 2 √ 2, 3 + 3 √ 2
Solution: Here, the common difference is;
Here, a2 – a1 = 3 + √ 2 − 3 = √ 2
a3 – a2 = ( 3 + 2 √ 2) − ( 3 + √ 2) = √ 2
a4 – a3 = ( 3 + 3 √ 2) – ( 3 + 2 √ 2) = √ 2
Since, an + 1 – an or the common difference is same every time.
Therefore, d = √ 2 and the given series forms an A.P.
Hence, next three terms are;
a5 = ( 3 + √ 2) + √ 2 = 3 + 4 √ 2
a6 = ( 3 + 4 √ 2) + √ 2 = 3 + 5 √ 2
a7 = ( 3 + 5 √ 2) + √ 2 = 3 + 6 √ 2
( vi) 0.2, 0.22, 0.222, 0.2222…
Solution: Here, the common difference is;
a2 – a1 = 0.22 − 0.2 = −0.02
a3 – a2 = 0.222 − 0.22 = 0.002
∴ 0.22 − 0.2 ≠ 0.222 − 0.22
Since the difference is not equal it does not form an A..P.
( vii) 0, − 4, − 8, − 12…
Solution: Here, the common difference is;
a2 – a1 = − 4 – 0 = −4 ,
a3 – a2 = − 8 − ( − 4) = − 4
a4 – a3 = − 12 − ( − 8) = − 4
∴ Common difference ( d) = − 4
Since, an + 1 – an or the common difference is same every time.
Fifth term = − 12 – 4 = − 16 Sixth term = − 16 – 4 = − 20
Seventh term = − 20 – 4 = − 24
Therefore, next three terms are − 16, − 20 and − 24
( viii) − 1 / 2, − 1 / 2, − 1 / 2, − 1 / 2 ….
Solution:Here, the common difference is;
Here, a2 – a1 = ( − 1 / 2) – ( − 1 / 2) = 0
a3 – a2 = ( − 1 / 2) – ( − 1 / 2) = 0
a4 – a3 = ( − 1 / 2) – ( − 1 / 2) = 0
Since, an + 1 – an or the common difference is same every time.
Therefore, d = 0 and the given series forms a A.P.
Hence, next three terms are;
a5 = ( − 1 / 2) − 0 = − 1 / 2
a6 = ( − 1 / 2) − 0 = − 1 / 2
a7 = ( − 1 / 2) − 0 = − 1 / 2
( ix) 1, 3, 9, 27…
Solution: Here, the common difference is;
a2 – a1 = 3 – 1 = 2
a3 – a2 = 9− 3 = 6.
Since, an + 1 – an or the common difference is not same every time.
Therefore, the given series doesn’t forms a A.P.
( x) a, 2a, 3a, 4a…
Solution: Here, the common difference is;
2a – a = 3a − 2a = 4a − 3a = a
Common difference ( d) = a
Fifth term = 4a + a = 5a Sixth term = 5a + a = 6a
Seventh term = 6a + a = 7a
Therefore, next three terms are 5a, 6a and 7a
( xi) a, a2, a3, a4 …
Solution: Here, the common difference is;
Here, a2 – a1 = a2–a = a ( a − 1)
a3 – a2 = a3 – a2 = a2 ( a − 1)
a4 – a3 = a4 – a3 = a3 ( a − 1)
Since, an + 1 – an or the common difference is not same every time.
Therefore, the given series doesn’t forms a A.P.
( xii) √ 2, √ 8, √ 18, √ 32 …
Solution: Here, a2 – a1 = √ 8 − √ 2 = 2 √ 2 − √ 2 = √ 2
a3 – a2 = √ 18 − √ 8 = 3 √ 2 − 2 √ 2 = √ 2
a4 – a3 = 4 √ 2 − 3 √ 2 = √ 2
Since, an + 1 – an or the common difference is same every time.
Therefore, d = √ 2 and the given series forms a A.P.
Hence, next three terms are;
a5 = √ 32 + √ 2 = 4 √ 2 + √ 2 = 5 √ 2 = √ 50
a6 = 5 √ 2 + √ 2 = 6 √ 2 = √ 72
a7 = 6 √ 2 + √ 2 = 7 √ 2 = √ 98
( xiii) √ 3, √ 6, √ 9, √ 12 …
Solution: Here, the common difference is;
a2 – a1 = √ 6 − √ 3 = √ 3× √ 2 − √ 3 = √ 3 ( √ 2 − 1)
a3 – a2 = √ 9 − √ 6 = 3 − √ 6 = √ 3 ( √ 3 − √ 2)
a4 – a3 = √ 12 – √ 9 = 2 √ 3 – √ 3× √ 3 = √ 3 ( 2 − √ 3)
Since, an + 1 – an or the common difference is not same every time.
Therefore, the given series doesn’t form a A.P.
( xiv)12, 32, 52, 72 … Or, 1, 9, 25, 49 …..
Solution: Here, the common difference is;
Here, a2 − a1 = 9 − 1 = 8
a3 − a2 = 25 − 9 = 16
a4 − a3 = 49 − 25 = 24
Since, an + 1 – an or the common difference is not same every time.
Therefore, the given series doesn’t form a A.P.
( xv) 12, 52, 72, 73 … Or 1, 25, 49, 73 …
Solution: Here, the common difference is;
a2 − a1 = 25 − 1 = 24
a3 − a2 = 49 − 25 = 24
a4 − a3 = 73 − 49 = 24
Since, an + 1 – an or the common difference is same every time.
Therefore, d = 24 and the given series forms a A.P.
Hence, next three terms are;
a5 = 73 + 24 = 97
a6 = 97 + 24 = 121
a7 = 121 + 24 = 145
Additional Questions
1.In which of the following situations, does the list of numbers involved make an arithmetic progression, and why:
( a)The amount of money in the account every year, when Rs 15000 is deposited at compound interest at 9 % per annum.
- Write first four terms of the AP,when the first term a and the common difference d are given as follows: a = 12, d = 6
- For the following AP’s, write the first term and the common difference:
− 7, − 4, − 1, 2, 5, 8…………………………………………………………………
- Which of the following are APs? If they form an AP,find the common difference d and write three more terms : 3 ,6, 9, 12………………………
Finite or Infinite Arithmetic Progressions
- Finite Arithmetic Progression
If there are only a limited number of terms in the sequence then it is known as finite Arithmetic Progression.
229, 329, 429, 529, 629
- Infinite Arithmetic Progression
If there are an infinite number of terms in the sequence then it is known as infinite Arithmetic Progression.
2, 4, 6, 8, 10, 12, 14, 16, 18…..…
The nth term of an Arithmetic Progression
If an is the nth term, a1 is the first term, n is the number of terms in the sequence and d is a common difference then the nth term of an Arithmetic Progression will be
Arithmetic Series:
The arithmetic series is the sum of all the terms of the arithmetic sequence.
The arithmetic series is in the form of
{a + ( a + d ) + ( a + 2d ) + ( a + 3d ) + ………}