Arithmetic Progression
Question 1. The nth term of a Sequence In defined as follows. Find the first four terms:
1. an=3n+1
Solution:
Given: an=3n+1
3n=3nt
Put n= 1,2,3,4, we get
a1 = 3×1+1=4
a2 = 3×2+1=7
a3 = 3×3 +1 = 10
a4=3×4+1= 13
The first four terms of the Sequence are 4,7, 10,13.
2. an= n2+3
Solution:
Given:
an=n2+3
Put n=1,2,3,4 we get
a1= (1)2+3 = 4
a2=(2)2 +3= 7
a3 = (3)2+3 = 12
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a4=(4)2 +3= 19
The first four terms of the Sequence are 4, 7, 12, and 19.
3. an = n(n+1)
Solution:
an=n (n+1)
Put n= 1,2,3,4 we get
a1 = 1(1+1)=2
a2 = 2(2+1)=6
a3=3(3+1)= (2
a4 = 4(4+1)=20
The first four terms of the Sequence are 2,6,12,20
4.\(a_n=n+\frac{1}{n}\)
Solution:
Given:
⇒ \(a_n=n+\frac{1}{n}\)
Put n=1,2,3,4 we get
⇒ \(a_1=1+\frac{1}{1}=2 \)
⇒ \(a_2=2+\frac{1}{2}=\frac{4+1}{2}=\frac{5}{2} \)
⇒ \(a_3=3+\frac{1}{3}=\frac{9+1}{3}=\frac{10}{3} \)
⇒ \(a_4=4+\frac{1}{4}=\frac{16+1}{3}=\frac{17}{3}\)
First four terms of the Sequence are 2,\( \frac{5}{2}, \frac{10}{3}, \frac{17}{3}\)
5. an= 3n
Solution:
Given
an=3n
Put n=1,2,3,4 we get
a1=31=3
a2=32 = 9
a3=33=27
a4=34=81
The first four terms of the Sequence are 3, 9, 27, 81
Question 2. The nth term of a Sequence is (3n-7). Find its 20th term.
Solution:
3n-7
n=20
3(20)-7
60-7
57
The 20th term of a Sequence is 57
Question 3. Which of the following are A.p.’s? If they form an A.P., find the Common difference ‘d’ and write three more terms:
1. -10, -6, -2, 2,…….
Solution:
Here a=-10,
d=-6-10=4
-10, -6, -2, 2,4,6,10,14
Yes d= 4, next three items = 6, 10, 14
2. 3,3+ √2, 3+252, 3+3√2,…..
Solution:
Here a=3
d=3+√2-3 =) √2
3, 3+ √2, 3+2√2, 3+3√2, 3+4√2,3+5√2
Yes d=52, next three terms = 3+3√2, 3+4√2,3+5√2.
3. 0, -4, -8,-12,
Solution:
Here a=0
d=0-4=-4
0,-4,-8,-12-16,-20,-24
Yes do, next three terms = -16,-20,-24
Question 4. For the following A.-P. ‘S write the first terms and Common differences:
1. 2,5, 8, 11,..
Solution:
2, 5, 8, 11,
Her first term a=2
Common difference = 5-2 = 3
2. -5,-1, 3, 7,
Solution: -5,-1, 3, 7
Her first term a=-5
Common difference = -541 = 4
Question 5. write the first four terms of the Ap., when the first term ‘a’ and the Common difference ‘d’ are given as follows:
1. a=5, d=3
Solution:
a=5, d= 3
a1 =5
a2=5+3=8
a3=8+3=11
a4=11+3= 14
The first four terms are 5, 8, 11, 14
2. a=-2,d=4
Solution:
a=-2, d=4
a1 =-2
a2=-2+4=2
a3=2+4=6
a4=6+4= 10
The first four terms are -2,2,6,10
Question 6. Find the 10th term of the AP: 1,3,5,7,..
Solution:
Here, a=1
d=3-1=2,
n = 10
an = a+ (n-1)
90= 1+ (10-1)2
⇒96=1+18
⇒ 210=19
10th term of the given Ap=19
Question 7. Find the 7th term of the AP 80, 77,74,71,
Solution:
Here a=80
d=77-80=-3,
n=7
a7 = a + (n-1)d
a7=80+(7-1)-3
a7=80-18
⇒ a7 = 62
7th term of the given A.p. =62
Question 8. Find the nth term of the A⋅p: -5, -3, -1, 1, —-
Solution:
Here a = -5
d=-3-5=2
n = n
an = a+ (n-1) d
an=-5+ (n-1)2
⇒ an=-5+2n-2
⇒ an= 2n-7
nth term of the given A.p. = (2n-7)
Question 9. Which term of the A-P. 4, 8, 12, is 76?
Solution:
Here, a = 4,
d=8-4=4
Let an=76
=) 4+ (n-1)4=76
= 4+4n-4=76
4n=76
⇒ \(n=\frac{76}{4} \Rightarrow n=19\)
19thterm of the given A.P. is 76
Question 10. which term of the Ap. 36, 33, 30, is Zero?
Solution:
Here a=36
d=33-36=-3
let an = 0
36+ (n-1)-3=0
36-3n+3=0
-3n=-39
\(n=\frac{39}{3} \Rightarrow n=13\)The 13th term of the given A.P. is zero
Question 11. which term of the\(\frac{3}{4}, 1, \frac{5}{4}, \ldots \text {. is } 12 ?\)
Solution:
⇒ \(\text { Here } a=\frac{3}{4} \)
⇒ \(d=1-\frac{3}{4}=\frac{1}{4} \)
⇒ \(\text { let } a_n=12 \)
⇒ \(\frac{3}{4}+(n-1) \frac{1}{4}=12 \)
⇒ \(\frac{3+(n-1)}{4}=12\)
3+n-1=48
⇒ n+2=48
⇒n=46
46th term of the given AP is 12.
Question 12. Find the number of terms in the Ap. 8, 12, 16,
Solution:
Here a=8
d=12-8=) 4
let an=124
=) 8+ (n-1)4=124
=) 8+40-4=124
=) 4n=124-4
4n = 120
⇒ n = \(n=\frac{120}{4} \Rightarrow 30\)
30th term of the given A.P. is 124.
Question 13. Find the number of terms in the A.p. 75, 70, 65, 15
Solution:
Here a=75
d=70-75=-5
let an=15
75+ (n-1)-5=15
75-5n+5=15
70-5n=15
-517=15-70
-5n=-55 ⇒ n=11
11th term of the given A. p. is 15.
Question 14. Find the 10th term from the end of the A.P. 82, 79, 76, —-,4.
Solution:
Here aa 1=4
d=79-82=-3
n=10
(-(10-1)defined
– 4-(10-1)-3
=4727
=31
10th term from the end = 31
Question 15. Find the 16th term from the end of the A.P. 3,6,9,99
Solution:
Here, 1=99
d= 6-3 = 3,
n=16
⇒ 99-(16-1) 3
⇒ 99-45
⇒ 54
16th term from the end = 54
Question 16. Find the Sum of the following A.p.: 3,8, 13, to 20 terms
Solution:
S1 =3+8+13,
a1=3
d=8-3=5
n=20
⇒ \(S_n=\frac{n}{2}[2 a+(n-1) d]] \)
⇒ \(S_{20}=\frac{20}{2}[2(3)+(20-1) 5]\)
⇒ \(S_{20}=10[6+95]\)
⇒ \(S_{20}=10[101]\)
⇒ \(S_{20}=1010 \)
Question 17. Find the sum of the following A.p. 5: 1,4,7,– to 50 terms
Solution:
S = 1+4+7+—–50
a=1
d=4-1= 3
n=50
⇒ \(S_n=\frac{n}{2}[2 a+(n-1) d] \)
⇒ \(S_{50}=\frac{50}{2}[2(1)+(50-1) 3] \)
⇒ \(S_{50}=25[2+147] \)
⇒ \(S_{50}=25[(49]\)
⇒ \(S_{50}=3,725\)
Question 18. Find the Sum given below: 3+6+9+ …..+96
Solution:
S=3+6+9+…… +96
a1 =3, d= 6-3 =) 3
an=96
a1+ (n-1) d=96
3+(n-1)3=96
3+3n-3=96
3n=96
⇒ \(n=\frac{96}{3} \Rightarrow n=32\)
S1 = Sum of 32 terms with first 3 terms and last term 96
⇒ \(S_1=\frac{32}{2}[3+96]\)
⇒ \(S_1=\frac{32}{2}[99]\)
⇒ \(S_1=16[99]\)
⇒ \(S_1=1584\)
Question 19. Find the Sum given below! 2 + 4+ 6+.
Solution:
S=2+4+6+—–+50
a1 =2,
d=4-2 =) 2 Q1=2,
an=50
a1+ (n-1)d=50
2+(n-1)2=50
2+2n=2=50
⇒ \(n=\frac{50}{2} \Rightarrow n=25\)
S1 = Sum of 25 terms with first 2 terms and last terms so
⇒ \(S_1=\frac{50}{2}[2+50] \)
⇒ \(S_1=\frac{50}{2}[526] \)
⇒ \(S_1=50[26] \)
⇒ \(S_1=650\)
Question 20. In an A.p.: given a=2, d= 3, Qn = 50, find n and Sn.
Solution:
Given a=2, d=3, an =50
a+(n-1)d=an
2+(n-1)3=50
2+3n-3=50
3n=5041
⇒ \(n=\frac{51}{3} \Rightarrow n=17 \)
⇒ \(S_n=\frac{n}{2}[2 a+(n-1) d] \)
⇒ \(5_n=\frac{17}{2}[2(2)+(17-1) 3] \)
⇒ \(S_n=\frac{17}{2}[2(2)+(17-1) 3] \)
⇒ \(S_n=\frac{17}{2}[4+48] \)
⇒ \(S_n=\frac{17}{2}\left[S_2\right] \)
⇒ \(S_n=17[26] \)
⇒ \(S_n=442\)
Question 21. Find the Value of x for which (x+2), 2x, (2x+3) are three consecutive terms of A.P.
Solution:
(x+2), 2x, (2x+3) are three consecutive terms of A.p.
\(2 x=\frac{(x+2)+(2 x+3)}{2}\)4x= x+2+2x+3
4x=3x+5
4x-3x=5
x=5