Arithmetic Progression II

Let the numbers be a-d, a, a+d
Given a-d + a + a+d = 21
3a = 21
a=7
Also, (a-d)a(a+d) = 280
a(a^2 – d^2) = 280
49 – d^2 = 40
d = sqrt (9) = 3
The 3 terms are 4, 7 and 10

Given
a+(3-1)d + a + (5-1)d = 30
i.e. 2a + 6d = 30—–(1)
a+(2-1)d + a + (7-1)d = 60
2a+7d=42——(2)
Solving linear equations (1) and (2)
d = 12
a = 15 – 3d = -21
The first three terms of the AP are -21, -9, 3

Between 100 and 200, the first and the last multiples of 3 are 102 and 198. This form an A.P. with
a = 102, l = 198, d = 3. We need to find n.
In a finite A.P. whose first term is a and last term l, then the number of terms in the A.P. is
given by n = (l – a)/d + 1
n = (198 – 102)/3 + 1 = 33

11th term is 38
nth term of an A.P. is
tn = a + (n-1)d
38 = a + (11-1) d
Also, 73 = a + (16-1) d
Solving both the equations,
a + 10d = 38 ——–(1)
a + 15d = 73 ——–(2)
Equation (2) – (1) gives
5d = 35
d = 7
a = 38 – 10d = 38 – 70 = -32
t21 = – 32 + (21-1)7 = 108

Given
(a+(3-1)d)/ a = 4/1
3a = 2d
d = 3a/2
ratio of 7th term to 2nd term is
(a+(7-1)d)/ (a+(2-1)d)
= (a+6d)/(a+d)
= (a+6 *3a/2)/(a+ 3a/2)
=10a * 2/5a = 4/1

Side length of the square garden = 80 /4 = 20 = a
d = 5
n = 4
nth term of an A.P. is
tn = a + (n-1)d
t4 = 20 + (4-1)5 = 35
Side length of the square garden after 4 years = 35 meters
Perimeter after 4 years = 4*35 = 140 meters

Let us take the four terms in the form (a-3d), (a-d), (a+d) and (a+3d) where the common difference is 2d
Given, a-3d = 10 —-(1)
(a-3d) + (a-d) + (a+d) + (a+3d) = 130
4a = 130
a = 22.5 ——-(2)
Substituting (2) in (1)
22.5 – 3d = 10
d = 7.5
Common difference = 2d = 15

Given a = Rs. 400, d = 150, n=5
The corresponding A.P. is
400, 550, 700, 850, 1000,…
Total rental cost for 5 days
= 400 + 550 + 700 + 850 + 1000
= 3500

Given a = 2,00,000
d = 5000, n=10
nth term of an A.P. is
tn = a + (n-1)d
t10 = 2,00,000 + (10-1) 5000
=2,45,000

Given a = 10, d = 2, n=8
nth term of an A.P. is
tn = a + (n-1)d
t8 = 10 + (8-1) 2 = 24 miles