Arithmetic Progression Examples.

Q: A man starts repaying a loan, with Rs 100 being its first instalment. If he increases the instalment by Rs 5/- every month, what amount he will pay in the 30^{th} instalment?

Solution:

We know that the last term of an AP is given by:

l = (a+ (n-1) * d). ….. (1)

Now, as per the question,

a = 100,d = 5 and n = 30.

- l = (100 + (30-1) * 5) substituting the values in expression (1)
- l = 245

Thus, the amount to be paid in the 30^{th} instalment will be Rs 245/-

***************************************************************

Q2. The Sum of the first four terms of an AP is 56, and, the sum of the last 4

terms is 112. If its first term is 11 then find out the number of terms of the

series.

SOLUTION.

We know that the Sum of the first n terms of an AP is given by:

Sn = (n / 2) * ((2 * a + (n-1) * d)). ……. (1)

Now, as per the question, substituting the relevant values in (1), we have,

- = (4 / 2) * ((2 * 11 + (4-1) * d)).
- 56 = 44 + 6d
- d = 2

Thus, now we have the first term as 11 and the common difference as 2.

Now, the last 4 terms of a AP is: a + (n-1) * d, a + (n-2) * d , a + (n-3) * d and a + (n-4) * d.

Now, as per the question,

a + (n-1) * d + a + (n-2) * d + a + (n-3) * d + a + (n-4) * d = 112

- 4a+4nd – 10d = 112
- 2d(2n – 5) + 4 * 11 = 112 (substituting d = 2,a = 11)
- 4(2n-5) + 4 * 11 = 112
- 8n-20+44 = 112
- 8n = 112 – 24 = 88
- n = 11

Thus, the number of the terms of the series is 11.