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Arithmetic Progressions

    nth Term of an AP

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10th class > Arithmetic Progressions > nth Term of an AP

nth Term of an AP

Let us consider the situation again, given in the earlier section in which Reena applied for a job and got selected. She has been offered the job with a starting monthly salary of ₹ 8000, with an annual increment of ₹ 500. What would be her monthly salary for the fifth year?

To answer this, let us first see what her monthly salary for the second year would be.

It would be ₹ (8000 + 500) = Rs. .

In the same way, we can find the monthly salary for the 3rd, 4th and 5th year by adding ₹ 500 to the salary of the previous year.

So, the salary for the 3rd year = Rs. (8500 + 500) = ₹ (8000 + 500 + 500) = ₹ (8000 + 2 × 500) = ₹ [8000 + (3 – 1) × 500] (for the 3rd year) = ₹

Salary for the 4th year = ₹ (9000 + 500) = ₹ (8000 + 500 + 500 + 500) = ₹ (8000 + 3 × 500) = ₹ [8000 + (4 – 1) × 500] (for the 4th year) = ₹

Salary for the 5th year = ₹ (9500 + 500) = ₹ (8000+500+500+500 + 500) = ₹ (8000 + 4 × 500) = ₹ [8000 + (5 – 1) × 500] (for the 5th year) = ₹

Observe that we are getting a list of numbers 8000, 8500, 9000, 9500, 10000, . . .

These numbers are in AP with a = and d = .

Now, looking at the pattern formed above, can you find her monthly salary for the 6th year? The 15th year? And, assuming that she will still be working in the job, what about the monthly salary for the 25th year? We would calculate this by adding ₹ 500 each time to the salary of the previous year to give the answer.

Can we make this process shorter? Let us see. You may have already got some idea from the way we have obtained the salaries above.

Salary for the 15th year = Salary for the 14th year + ₹ 500

= ₹ [8000 + (500 + 500 + ........ 13 times)] + 500

= ₹ [8000 + × 500]

= ₹ [8000 + (15 – 1) × 500] = ₹

i.e., First salary + (15 – 1) × Annual increment.

In the same way, her monthly salary for the 25th year would be ₹ [8000 + (25 – 1) × 500] = ₹

= First salary + (25 – 1) × Annual increment

This example would have given you some idea about how to write the 15th term, or the 25th term, and more generally, the nth term of the AP.

Let a1, a2, a13 , . . . be an AP whose first term a1 is a and the common difference is d.

Then, the second term a2 = a + d = a + (2 – 1) d

the third term a3 = a2 + d = (a + d) + d = a + 2d = a + (3 – 1) d

the fourth term a4 = a3 + d = (a + 2d) + d = a + 3d = a + (4 – 1) d

. . . . . . . . . . . . . . . .

Looking at the pattern, we can say that the nth term an = a + (n – 1) d

So, the nth term an of the AP with first term a and common difference d is given by

an = a + (n – 1) d.

an is also called the general term of the AP. If there are m terms in the AP, then am represents the last term which is sometimes also denoted by l.

Let us consider some examples.

Example 3 : Find the 10th term of the AP : 2, 7, 12, . . .

Find tenth term

  • Here, we have: a = while d =
  • Using the nth term formula: an=a+n−1d, we substitute the values to get the answer with n = .
  • Tenth term for given AP:
  • We have found the answer.

Example 4 :Which term of the AP : 21, 18, 15, . . . is – 81? Also, is any term 0? Give reason for your answer.

Find the value of n

  • Here, we have: a = while d = and an =
  • Using the nth term formula: an=a+n−1d, we substitute the values to get the answer.
  • Putting the values
  • The term −81 is the th term for the given AP.
  • Now, to find out if 0 is also part of the AP: We substitute an = 0 and see if upon solving the value of n is .
  • Substituting the values and solving
  • We get n =
  • Thus 0 is a part of the AP.

Example 5

Determine the AP whose 3rd term is 5 and the 7th term is 9.

Solution : We have:

a3 = a + (3 – 1) d = a + = (1)

and a7 = a + (7 – 1) d = a + = (2)

Since, we need to know the value of a and d to form an AP, we can solve the above equations to get the desired answer.

Using elimination method: We can subtract (1) from (2),

a + 6d = (2)

-a -2d = (1)

_________________

d = which further gives d =

Substituting d = 1 in equation (1), we get a =

as a + 2(1) = 5 gives a = 5−2 = .

Hence, the required AP is 3, , , , and so on.

Example 6

Example 6 : Check whether 301 is a term of the list of numbers 5, 11, 17, 23, . . .

Solution : We have :

a2−a1 = 11 – 5 = ,

a3−a2 = – = ,

a4−a3 = – =

Thus, the set of numbers form an AP.

Now, a = and d =

Let 301 be the nth term of this AP. Thus,

an=a+n−1d

301=5+n−1×6

301 = –

So, n = 3026 =

So, 301 is not a term of the given list of numbers as it needs to be a positive integers to prove that it is a part of the AP.

Example 7

Example 7 : How many two-digit numbers are divisible by 3?

Solution :

The list of two-digit numbers divisible by 3 is :

12, 15, 18, . . . , 99

Is this an AP?

Since, the common difference remains the same i.e.

Here, a = , d = , an =

As an=a+n−1d,

99 = 12 + (n – 1) × 3

= (n – 1) × 3

n =

So, there are 30 two-digit numbers divisible by 3.

Example 8

Example 8 : Find the 11th term from the last term (towards the first term) of the AP : 10, 7, 4, . . ., – 62.

Solution :

Here, a = 10, d = , l = – 62, where l = a + (n – 1) d

To find the 11th term from the last term, we will find the total number of terms in the AP.

So, = + (n – 1)

= (n – 1)(–3)

Solving, we get:

n - 1 =−72−3

n – 1 =

n =

So, there are 25 terms in the given AP.

The 11th term from the last term will be the th term. [Hint: No.of terms - 11(as mentioned 11th from the last term ) + 1]

So, a15 = 10 + (15 – 1)(–3) =

Thus, the 11th term from the last term is – 32.

Alternative Solution :

If we write the given AP in the reverse order, then a = – 62 and d =

So, the question now becomes finding the 11th term with these a and d.

So, a11 = – 62 + (11 – 1) × 3 = −32

So, the 11th term, which is now the required term, is −32 (same as before).

Example 9 : A sum of Rs. 1000 is invested at 8% simple interest per year. Calculate the interest at the end of each year. Do these interests form an AP? If so, find the interest at the end of 30 years making use of this fact.

Is it an AP?

  • Formula to calculate simple interest is given by Simple Interest = P×R×T100
  • We also have P = ₹ and R =
  • Substituting the values to get interest for 1st year, we get:₹
  • Substituting the values to get interest for 2nd year, we get: ₹
  • Substituting the values to get interest for 3rd year, we get: ₹
  • Thus, number list : 80, 160, 240 .....
  • On checking for common difference, we see that it an AP.
  • Thus, finding the interest for the 30th year, we get a30: ₹
  • We have found the answer.

Example 10 : In a flower bed, there are 23 rose plants in the first row, 21 in the second, 19 in the third, and so on. There are 5 rose plants in the last row. How many rows are there in the flower bed?

Find number of rows

  • We see that: 23, 21, 19, . . ., 5 an AP.
  • Let the number of rows in the flower bed be n. Then a = , d = , an =
  • Putting values in the formula: an = where a, d and n mean the same as we have studied in this section.
  • Putting values into the equation.
  • So, n =
  • Thus, there are a total of ten rows.