Arithmetic Progression-1
Arithmetic Progression (A. P.):
A sequence is called arithmetic Progression if the difference of a term and the previous term is always same.
Ex: 2,4,6 .... are in A. P.
If 'a' is the first term and 'd' is the common difference of an A. P. then the 'n' th term is given by
an =a+(n –1)d
Here the terms are a, a+d, a+2d,...... a+(n –1)d
Terms of A. P. represented by
a1, a2, a3.....,
or
t1, t2, t3.....,
Ex: 2,4,6 .... are in A. P.
If 'a' is the first term and 'd' is the common difference of an A. P. then the 'n' th term is given by
an =a+(n –1)d
Here the terms are a, a+d, a+2d,...... a+(n –1)d
Terms of A. P. represented by
a1, a2, a3.....,
or
t1, t2, t3.....,
One mark Questions :
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1) Find the sum of all 11 terms of an AP whose middle term is 30
Answer is 330
In an AP with 11 terms,
the middle term is (11+1)/2 = 6th term
⇒Now a6= a+5d =30
S11=11/2[2a+10d]
=11(a+5d) = 11× 30 =3302) If 4 times the 4th term of an AP is equal to 18 times the 18th term, then find the 22nd term.
Answer : The 22nd term of AP is '0'
Let 'a' be the 1st term and 'd' be the common difference of the AP
Now an=a+(n –1)d
Given that
4 ×a4=18×a18
4(a+3d) = 18(a+17d)
2a+6d=9a+153d
7a = –147d
a= –21d
a+21d=0
a+(22 –1)d=0
a22=0
3)If 18, a, b, –3 are in A.P, then find a+b
Answer : 15
If 18,a,b, –3 are in AP.
Then a –18= –3 –b [ ∵ Common difference is equal]
a+b = –3 +18
a+b =15
4)The sum of first 20 terms of the A.P 1, 4, 7,10..... is
Answer : 590
Given A.P is 1,4,7,10...
a=1, d=4 –1=3, n=20
Sn= (n/2)[2a+(n –1)d]
S20=(20/2)[2(1)+(20 –1)(3)]
=10[2+57]
=10(59)
=5905)What is the 'n'th term of the AP. a, 3a, 5a.....
Answer : (2n –1)a
Given AP is a,3a,5a...
t1=first term=a,
Common difference=d=3a –a=2a
n th term = t1+(n –1)d
=a+(n –1)2a
=a+2an –2a
=2na –a
=(2n –1)a6)In the AP 2, x, 26 find the value of 'x'
Answer :14
If a,b,c are in A.P.,
2b=a+c ( ∵ a, b, c are in A.P b –a =c –b)
Given 2, x, 26 are in A.P
2x=26+2
x=(28/2) =14
7)For what value of 'k' k+2, 4k–6, 3k–2 are three consecutive terms of an A.P
Answer :3
If a, b, c are consecutive terms of A.P.
Then 2b=a+c
If k+2, 4k –6, 3k –2 are in A.P.
⇒2(4k –6)=k+2+3k –2
8k –12=4k
8k –4k=12
⇒ 4k=12
⇒ k=3
8) What is the common difference of an A.P which a21 – a7 =84
Answer is 6
a21 – a7 =84
(a+20d) –(a+6d)=84
a+20d – a –6d =84
14d =84
⇒ d =(84/14)=6
Common difference =6
9)If the common difference of an A.P., is –6 find a16 – a12
Answer : –24
Let the first term is 'a' and common difference is 'd'
Given d = –6
a16 – a12= (a+15d) – (a+11d)
= a+15d –a –11d =4d
= 4( –6)= –24
10) For what value of 'k' will the consecutive terms 2k+1, 3k+3 and 5k –1 form an A.P.?
Answer is 6
we say a, b, c are in A.P. if 2b =a+c
Thus if 2k+1, 3k+3, 5k –1 are in A.P.
If 2(3k+3)=(2k+1)+(5k –1)
6k+6=7k
⇒ 6=7k –6k
k=6
For k=6 given consecutive terms form an A.P.
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1) Find the sum of all 11 terms of an AP whose middle term is 30
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Answer is 330 In an AP with 11 terms,
the middle term is (11+1)/2 = 6th term
⇒Now a6= a+5d =30
S11=11/2[2a+10d]
=11(a+5d) = 11× 30 =330
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Answer : The 22nd term of AP is '0' Let 'a' be the 1st term and 'd' be the common difference of the AP
Now an=a+(n –1)d
Given that
4 ×a4=18×a18
4(a+3d) = 18(a+17d)
2a+6d=9a+153d
7a = –147d
a= –21d
a+21d=0
a+(22 –1)d=0
a22=0
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Answer : 15 If 18,a,b, –3 are in AP.
Then a –18= –3 –b [ ∵ Common difference is equal]
a+b = –3 +18
a+b =15
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Answer : 590 Given A.P is 1,4,7,10...
a=1, d=4 –1=3, n=20
Sn= (n/2)[2a+(n –1)d]
S20=(20/2)[2(1)+(20 –1)(3)]
=10[2+57]
=10(59)
=590
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Answer : (2n –1)a Given AP is a,3a,5a...
t1=first term=a,
Common difference=d=3a –a=2a
n th term = t1+(n –1)d
=a+(n –1)2a
=a+2an –2a
=2na –a
=(2n –1)a
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Answer :14 If a,b,c are in A.P.,
2b=a+c ( ∵ a, b, c are in A.P b –a =c –b)
Given 2, x, 26 are in A.P
2x=26+2
x=(28/2) =14
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Answer :3 If a, b, c are consecutive terms of A.P.
Then 2b=a+c
If k+2, 4k –6, 3k –2 are in A.P.
⇒2(4k –6)=k+2+3k –2
8k –12=4k
8k –4k=12
⇒ 4k=12
⇒ k=3
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Answer is 6 a21 – a7 =84
(a+20d) –(a+6d)=84
a+20d – a –6d =84
14d =84
⇒ d =(84/14)=6
Common difference =6
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Answer : –24 Let the first term is 'a' and common difference is 'd'
Given d = –6
a16 – a12= (a+15d) – (a+11d)
= a+15d –a –11d =4d
= 4( –6)= –24
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Answer is 6 we say a, b, c are in A.P. if 2b =a+c
Thus if 2k+1, 3k+3, 5k –1 are in A.P.
If 2(3k+3)=(2k+1)+(5k –1)
6k+6=7k
⇒ 6=7k –6k
k=6
For k=6 given consecutive terms form an A.P.
