2.Nth Term of Arithmetic Progression

Nth Term Of An Arithmetic Progression

We know what an arithmetic progression (A.P.) is. Also, we have learnt that there is a common difference between any two consecutive terms of an A.P..

Now, can we find the required term of a given A.P. with this information? Let us consider the A.P. 3, 7, 11, 15,...

Here, first term (a) = 3 and common difference (d) = 4

Now, if we want to find the 5th term of this A.P., then we will simply add the common difference to 4th term. Thus, 5th term of this A.P. will be 15 + 4 = 19.

What would we do if we are asked to find the 20th term or 100th term or nth term? Obviously, the process of adding common difference will be very time consuming.

For such problems, we must have a short cut or a formula to find the general term of an A.P. Let us derive the same.

Consider the A.P. a, a + d, a + 2d, a + 3d, ...

For this A.P., we have

a₁ = a

a₂ – a₁ = d a₃ – a₂ = d

. . .

. . .

. . .

a_n – 1 – a_n – 2 = d

an – an – 1 = d

Adding all these equations, we get

a₁ + (a₂ – a₁) + (a₃ – a₂) +...+ (a_n – 1 – a_n – 2) + (a_n – a_n – 1) = a + {d + d +... + d (n – 1 times)}

⇒(a₁ – a₁) + (a₂ – a₂) + (a₃ – a₃) +...+ (a_n – 1 – a_n – 1) + a_n = a + (n – 1)d

⇒a_n = a + (n – 1)d

Hence, the general term or nth term i.e., an of an A.P. whose first term is a and common difference is d can be found by the following formula:

a_n = a + (n – 1)d

Sometimes, we need to find three, four or five consecutive terms of an A.P. then it is convenient to take them as follows:

• Three consecutive terms can be taken as a – d, a, a + d
• Four consecutive terms can be taken as a – 3d, a – d, a + d, a + 3d. Here, common difference is 2d.
• Five consecutive terms can be taken as a – 2d, a – d, a, a + d, a + 2d

Result:

In an A.P., common difference, , where Tp and Tq are pth and qth term respectively.

In particular, .

Proof:

T_p = a + (p – 1)d, T_q = a + (q – 1)d

⇒T_p – T_q = a + (p – 1)d – {a + (q – 1)d} = (p – q)d

In particular, take T₁ = a.

Then,

Now, let us solve some examples to understand the concept better.

Example 1: Find the 20th term of the following arithmetic progression. 0.4, 1.5, 2.6, 3.7, 4.8 …

Solution:

Here, a = 0.4 and d = 1.5 – 0.4 = 1.1

Thus, the 20^th term is given by,

a₂₀ = a + (20 – 1) d

= 0.4 + (20 – 1) 1.1

= 0.4 + 19 × 1.1

= 0.4 + 20.9

= 21.3

Thus, the 20th term of the given A.P. is 21.3.

Example 2: If the 7^th term of an A.P. is – 21 and 15th term is – 53, then find the first term and common difference.

Solution:

Let the first term and common difference of the A.P. be a and d respectively.

It is given that a7 = –21 and a₁₅ = –53

Using the formula for nth term, we obtain

a₇ = a + (7 – 1) d

– 21 = a + 6d … (1)

and, a₁₅ = a + (15 – 1) d

– 53 = a + 14d … (2)

Subtracting equation (1) from (2), we obtain

–32 = 8d

⇒ d = –4

Substituting the value of d in equation (1), we obtain

–21 = a + 6 (–4)

–21 = a – 24

⇒ a = 3

Thus, the first term is 3 and the common difference is –4.

Example 3: Is 102 a term of the A.P., 5, 11, 17, 23 …?

Solution:

Let 102 be the nth term of the given sequence.

∴ a_n = 102

Using the formula for nth term, we obtain

a_n= a + (n – 1) d

∴ 102 = a + (n – 1) d

For the given A.P., a = 5 and d = 11 – 5 = 6

∴ 102 = 5 + (n – 1) 6

102 – 5 = (n – 1) 6

97 = (n – 1) 6

However, n should be a positive integer. Therefore, 102 is not a term of the given A.P.

Example 4: Find the number of three-digit numbers that are divisible by 5.

Solution:

The first three-digit number which is divisible by 5 is 100, second is 105, third is 110, and so on. The last three-digit number which is divisible by 5 is 995.

Thus, we obtain the following A.P. 100, 105 … 995

Here, we have to find the number of terms, n.

∴Last term of A.P. = 995

The number of terms in the A.P. is n, so the last term is the nth term.

a + (n – 1)d = 995

Here, a = 100 and d = 5

100 + (n – 1)5 = 995

(n – 1)5 = 995 – 100

5n – 5 = 895

5n = 895 + 5

5n = 900

n = 180

Thus, there are 180 three-digit numbers, which are divisible by 5.

Example 5: The fare of a bus is Rs 10 for the first kilometre and Rs 5 for each additional kilometre. Find the fair after 12 kilometres.

Solution:

The fare after each kilometre forms an A.P. as follows.

Fare after one kilometre = Rs 10

Fare after two kilometres = 10 + 5 = Rs 15

Fare after three kilometres = 15 + 5 = Rs 20

Now the arithmetic progression is 10, 15, 20 …

Here, first term, a = 10 and common difference, d = 5

Now the fare after 12 kilometres is the 12th term of the A.P.

∴ a₁₂ = a + (12 – 1) d

a₁₂ = 10 + 11 × 5

= 10 + 55

= 65

Thus, the fare after 12 kilometres is Rs 65.

Example 6: Mohit borrowed a sum of money at a simple interest rate of 2% per annum. He has to pay an amount of Rs 1120 after 6 years. How much money did he borrow?

Solution:

Let the amount of money Mohit borrowed be Rs x. We know that the amount after T years is

Where, P and R denotes the principal and rate respectively

The amount after every year forms an A.P.

Amount after first year

Amount after second year

Thus, the A.P. is as follows.

Here, the first term is  and common difference is .

Now, it is given that the amount after 6 years is Rs 1120 i.e., 6th term of the A.P. is 1120. Now using the formula, an = a + (n – 1)d, we obtain

1120 × 100 = 112x

x = 1000

Thus, Mohit borrowed Rs 1000.

Example 7: Find four consecutive terms of an A.P. such that the difference of the middle terms is 8 and the product of the extreme terms is 217.

Solution:

Let four consecutive terms of required A.P. be a – 3d, a – d, a + d, a + 3d.

According to the question, we have

a + d – (a – d) = 8

⇒ a + d – a + d = 8

⇒ 2d = 8

⇒ d = 4 Also,

(a – 3d)(a + 3d) = 217

⇒ a² – (3d)² = 217

⇒ a² – 9d² = 217

⇒ a2 – 9(4²) = 217

⇒ a² – 144 = 217

⇒ a² = 361

⇒ a = ±19

When d = 4 and a = 19, then four consecutive terms are:

19 – 3(4), (19 – 4), (19 + 4), 19 + 3(4)

i.e., 7, 15, 23, 31

When d = 4 and a = –19, then four consecutive terms are:

–19 – 3(4), (–19 – 4), (–19 + 4), –19 + 3(4)

i.e., –31, –23, –15, –7

Example 8: Find five consecutive terms of an A.P. such that the product of the extreme terms is –63 and product of second and fourth terms is –15.

Solution:

Let five consecutive terms of required A.P. be a – 2d, a – d, a ,a + d, a + 2d.

According to the question, we have

(a – 2d)(a + 2d) = –63

⇒ a2 – (2d)² = –63

⇒ a² – 4d² = –63 ...(1)

Also,

(a – d)(a + d) = –15

⇒ a² – d² = –15 ...(2)

On subtracting (1) from (2), we get

3d² = 48

⇒ d² = 16

⇒ d = ±4

On substituting d² = 16 in (2), we get

a² – 16 = –15

⇒ a² = 1

⇒ a = ±1

When a = 1 and d = 4, A.P. will be –7, –3, 1, 5, 9

When a = 1 and d = –4, A.P. will be 9, 5, 1, –3, –7

When a = –1 and d = 4, A.P. will be –9, –5, –1, 3, 7

When a = –1 and d = –4, A.P. will be 7, 3, –1, –5, –9

Example 9: If the 25th and 35th terms of an arithmetic progression are 121 and 171 respectively, then find the common difference of the A.P.

Answer:

T₂₅ = 121, T₃₅ = 171

It is known that common difference, .

Example 10: In an A.P, show that .

Answer:

Therefore,