**Arithmetic Series Word Problems with Answers :**

Here we are going to see, some practical problems involving arithmetic series.

**Question 1 :**

A man repays a loan of 65,000 by paying 400 in the first month and then increasing the payment by 300 every month. How long will it take for him to clear the loan?

**Solution :**

For the first month he is paying = 400

payment of second month = 400 + 300 = 700

payment of 3rd month = 700 + 300 = 1000

400 + 700 + 1000, ................

loan amount = 65000

(n/2)[2a + (n - 1)d] = 65000

(n/2)[2(400) + (n - 1)300] = 65000

(n/2)[800 + 300n - 300] = 65000

(n/2)[500 + 300n] = 65000

(n/2)[5 + 3n] = 650

n[5 + 3n] = 1300

5n + 3n^{2} = 1300

3n^{2 }+ 5n - 1300 = 0

3n^{2} - 60n + 65n - 1300 = 0

3n(n - 20) + 65(n - 20) = 0

(n - 20) (3n + 65) = 0

n = 20 or n = -65/3 (not acceptable)

Hence he will clear the loan amount in 20 months.

**Question 2 :**

A brick staircase has a total of 30 steps. The bottom step requires 100 bricks. Each successive step requires two bricks less than the previous step.

(i) How many bricks are required for the top most step?

(ii) How many bricks are required to build the stair case?

**Solution :**

Let "l" be the number of bricks in the last step

Number of bricks in the 1st step (a) = 100

Number of bricks in the 2nd step = 100 - 2 = 98

d = 98 - 100 = -2

number of steps (n) = 30

(i)

t_{30} = a + 29d

= 100 + 29(-2)

= 100 - 58

= 42

Hence in the top most step, we will have 42 bricks.

(ii) How many bricks are required to build the stair case?

S_{n} = (n/2)[a + l]

= (30/2)[100 + 42]

= 15(142)

= 2130

**Question 3 :**

If S_{1, }S_{2}, S_{3},....Sm are the sums of n terms of m A.P.’s whose first terms are 1,2, 3,...m and whose common differences are 1, 3, 5,..., (2m -1) respectively, then show that S_{1} + S_{2} + S_{3} +............S_{m} = (mn/2)(1 + mn)

**Solution :**

n = [(l-a)/d] + 1

n = [((2mn - n + 1) - (1 + n))/2n] + 1

n = [(2mn - n + 1 - 1 - n)/2n] + 1

n = [(2mn - 2n)/2n] + 1

n = (m - 1) + 1

n = m

S_{1} + S_{2} + S_{3} +............S_{m}

Hence proved.

**Question 4 :**

Find the sum

**Solution :**

a = (a-b)/(a+b)

d = (3a-2b)/(a+b) - (a-b)/(a+b)

d = [3a - 2b -(a - b)]/(a + b)

d = [3a - 2b -a + b]/(a + b)

d = (2a - b) / (a + b)

Sn = (n/2)[2a + (n - 1)d]

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