Arithmetic Progression Examples

                                             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/-

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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.