# Problem Study (Arithmetic Progression)

1.    The Length of a perimeter of a hexagon is 36cm. The lengths the sides of the
hexagon are in arithmetic progression and the length of the longest side is five times
the length of the shortest side. Find the length of each side.

### Solution

Let the length of the sides of the hexagon in ascending order be u1, u2, u3, u4,
u5 and u6.
By the problem u1, u2, u3, u4, u5, u6 is an A.P.
Let u1 = a,
u2 = a + d
u3 = a + 2d
u4 = a + 3d
u5 = a + 4d
u6 = a + 5d
Perimeter of the hexagon = 36 (given)
Hence, 6a + 15d = 36
2a + 5d = 12 ............ (1)
length of longest side = 5 ( length of shortest side)
u6 = 5u1
a + 5d = 5a
4a - 5d = 0 .............(2)
Solving equation (1) and (2), a = 2 and d = = 1.6
u1 = a = 2 cm
u2 = a + d = 3.6 cm
u3 = a + 2d = 5.2 cm
u4 = a + 3d = 6.8 cm
u5 = a + 4d = 8.4 cm
u6 = a + 5d = 10 cm

2.   The sum of n terms of an arithmetic progression is given by the formula Sn = 2n2 + n.
Find (a) the first term, (b) the common difference and (c) the tenth term.

### Solution

Sn = 2n2 + n
u1 = S1 = 2(1)2 + (1) = 3
the first term = a = 3
u1 + u2 = S2 = 2(2)2 + (2) = 10
u2 = S2 - S1 = 7
the common difference = d = u2 - u1 = 4
un = a + (n - 1)d
u10 = a + 9d= 3 + 9(4) = 39

3.   During 1996 a company increased its sales of television sets at a constant rate of 200
sets per month. Thus the number of television sets sold in February was 200 more
than in January, the number of television sets sold in March was 200 more than in
February and this pattern continued month by month throughout the year. Given that
the company sold 38, 400 television sets in 1996, calculate the number of television
sets sold in (i) January,  (ii) December.

### Solution

Let the number of television sets sold in January = a
and those sold in December = u12.
∴ d = 200, n = 12 and S12 = 38, 400
S$\frac{n}{2}$  { 2a + (n - 1) d }
S12 = 38, 400
∴  $\frac{\mathrm{12}}{2}$  (2a + 11 × 200) = 38,400
2a + 2,200  = 6,400
a  =2100
u= a + (n - 1)d
u12  = a + 11d = 2,100 + 11 (200) = 4,300
Therefore there were 2,100 television sets sold in January and 4,300 sets in December.

4.   Find the sum of all numbers between 200 and 1,000 which are exactly divisible by 15.

### Solution

All numbers between 200 and 1,000 which are exactly divisible by 15 are
210, 225, 240, ... , 990.
Here the terms are in an A.P., with a = 210, d = 15 and l = u= 990.
Since  u= a + (n - 1)d,
a + (n - 1)d = 990
210 + (n - 1)(15) = 990
n = 53
S= $\frac{n}{2}$ (a + l)
S53  $\frac{53}{2}$ (210 + 990) = 31,800

Credit : Problems Supported by : Sayar Idea Zaw