Chapter 26.5 - Sum of n terms - Solved examples

In the previous section we completed the discussion on how to determine the sum of first n terms of an AP. We also saw some solved examples. In this section we will see a few more solved examples.

Solved example 26.28
Find the sum of the following APs
(i) 2, 7, 12, . . . , to 10 terms
Solution:
1. In the given AP, a = 2, d = 12 -7 = 5 and n = 10
2. We have: S = ⁿ⁄₂ [2a + (n-1)d] ⇒ S = ¹⁰⁄₂ × [2 × 2 + (10-1)×5]
= 5 × [4 + 45] = 5 × [49] = 5 × 49 = 245

(ii)  ¹⁄₁₅, ¹⁄₁₂, ¹⁄₁₀,, . . . , to 11 terms
Solution:
1. In the given AP, a = ¹⁄₁₅, d = ¹⁄₁₀ - ¹⁄₁₂ = ¹⁄₆₀ and n = 11
2. We have: S = ⁿ⁄₂ [2a + (n-1)d] ⇒ S = ¹¹⁄₂ × [2 × ¹⁄₁₅ + (11-1)×¹⁄₆₀]
= ¹¹⁄₂ × [²⁄₁₅ + 10 × ¹⁄₆₀] = ¹¹⁄₂ × [²⁄₁₅ + ¹⁄₆] = ¹¹⁄₂ × [¹⁸⁄₆₀] = ¹¹⁄₂ × [³⁄₁₀] = ³³⁄₂₀

Solved example 26.29
How many terms of the AP: 24, 21, 18, . . . must be taken so that, their sum is 78?
Solution:
1. In the given AP, a = 24, d = 18 -21 = -3 and S_n = 78
2. We have to find n
3. We have: S_n = ⁿ⁄₂ [2a + (n-1)d] ⇒ 78 = ⁿ⁄₂[2 × 24 + (n-1)×-3] ⇒ 78 = ⁿ⁄₂[48 - 3n + 3]
⇒ 78 = ⁿ⁄₂[51 - 3n] ⇒ 156 = n[51 - 3n] ⇒ 156 = 51n - 3n² ⇒ 3n² - 51n + 156 = 0
4. Solving this equation, we get n = 4 and n = 13
5. So, sum of the first 4 terms of the AP starting from 24 will be equal to 78
6. Sum of the first 13 terms starting from 24 will also be equal to 78
7. How can the two sums be equal to 78?
Ans: (i) Consider the terms starting from the fifth upto the thirteenth
(ii) The sum of those terms will be zero. Because, some of those terms are positive, and the others negative. Those terms will cancel out each other.
This is shown in a tabular form here.

Solved example 26.30
Find the sums given below:
(i) 7 + 10¹⁄₂ + 14 + . . . + 84
Solution:
1. The given series is an AP because, 14 - 10¹⁄₂ = 3¹⁄₂.
Also 10¹⁄₂ - 7 = 3¹⁄₂.
2. So, in the given AP, a = 7, d = 3¹⁄₂ and last term l = 84
3. To find the sum, we need to know the number of terms. That means, we need to know the 'n value' of the last term 84. We did such problems in a previous section. [See solved example 26.15 at the beginning of section 26.3]
(i) We have:  n^th term = a + (n-1)d ⇒ 84 = 7 + (n-1) × 3¹⁄₂ 
⇒  84 = 7 + ⁷ⁿ⁄₂ - ⁷⁄₂ ⇒ ⁷ⁿ⁄₂ = 84 - 7 + ⁷⁄₂ ⇒ ⁷ⁿ⁄₂ = ¹⁶¹⁄₂ ⇒ 7n = 161 ⇒ n = ¹⁶¹⁄₇ = 23
(ii) So 84 is the 23^rd term of the given AP.
4. Now we know the first term, last term and n. We can use eq.26.2(b) that we saw in the previous section.
S = ⁿ⁄₂ [a + l] ⇒ S = ²³⁄₂ [7 + 84] = ²³⁄₂ [91] = 1046¹⁄₂.

(ii) -5 + (-8) + (-11) + . . . + (-230)
Solution:
1. The given series is an AP because, -11 - (-8) = -11 + 8 = -3.
Also -8 - (-5) = -8 + 5 = -3.
2. So, in the given AP, a = -5, d = -3 and last term l = -230
3. To find the sum, we need to know the number of terms. That means, we need to know the 'n value' of the last term -230. We did such problems in a previous section. [See solved example 26.15 at the beginning of section 26.3]. Also see the previous problem.
(i) We have:  n^th term = a + (n-1)d ⇒ -230 = -5 + (n-1) × -3 ⇒ -230 = -5 -3n + 3 
⇒ 3n = 230 -5 + 3 ⇒ 3n = 228 ⇒ n = 76
(ii) So -230 is the 76^th term of the given AP.
4. Now we know the first term, last term and n. We can use eq.26.2(b) that we saw in the previous section.
S = ⁿ⁄₂ [a + l] ⇒ S = ⁷⁶⁄₂ [-5 + -230] = ⁷⁶⁄₂ [-235] = -8930 

Solved example 26.31
In an AP: Given a = 5, d = 3, a_n = 50, find n and S_n.
Solution:
1. In this problem, the n^th term a_n is given as 50. 
2. The sum from the first term 5 to the n^th term 50 is to be calculated. 50 is the last term.
3. To find the sum, we need to know the number of terms. That means, we need to know the 'n value' of the last term 50. We did such problems in a previous section. [See solved example 26.15 at the beginning of section 26.3]. Also see the previous problem. 
(i) We have:  n^th term = a + (n-1)d ⇒ 50 = 5 + (n-1) × 3 ⇒ 50 = 5 +3n - 3 
⇒ 3n = 50 -5 + 3 ⇒ 3n = 48 ⇒ n = 16
(ii) So 50 is the 16^th term of the given AP.
4. Now we know the first term, last term and n. We can use eq.26.2(b) that we saw in the previous section.
S = ⁿ⁄₂ [a + l] ⇒ S = ¹⁶⁄₂ [5 + 50] = 8 × 55 = 440

Solved example 26.32
Given a = 8, a_n  = 62, S_n = 210, find n and d.  
Solution:
1. We know the first term, last term and S_n. We can use eq.26.2(b) that we saw in the previous section.
S = ⁿ⁄₂ [a + l] ⇒ 210 = ⁿ⁄₂ [8 + 62] ⇒ 210 = ⁿ⁄₂ [70] ⇒ n = ²¹⁰⁄₃₅ = 6
• So the 6^th term is 62
2. We can use eq.26.1 which gives the position of any term:
• n^th term = a + (n-1)d ⇒ 62 = 8 + (6-1)d ⇒ 62 = 8 + 5d ⇒ 5d = 54 ⇒ d = ⁵⁴⁄₅.

Solved example 26.33
Given a₃ = 15, S₁₀ = 125, find d and a₁₀.
Solution:
1. S₁₀, which is the sum of first 10 terms is given as 125
• The first term a is not given. So we will use the basic eq.26.2 which gives the sum of n terms:
2. We have: S = ⁿ⁄₂ [2a + (n-1)d] ⇒125 = ¹⁰⁄₂ × [2a + (10-1)×d] 
⇒125 = 5 × [2a + 9d] ⇒25 = 2a + 9d
3. The third term a3 is given as 15. We can use eq.26.1 which gives the position of any term:
• n^th term = a + (n-1)d ⇒ 15 = a + (3-1)d ⇒ 15 = a + 2d
4. So we have two equations:
(i) 2a + 9d = 25 
(ii) a + 2d = 15
• From (ii) we get: a = 15-2d
• Substituting this in (i) we get: 2 × (15-2d) + 9d = 25 ⇒ 30 -4d +9d = 25 
⇒ 30 + 5d = 25 ⇒ 5d = -5 ⇒ d = -1
• Substituting this value of d in (ii) we get: a + 2 ×-1 = 15 ⇒ a -2 = 15 ⇒ a = 17  
5. Now we can find a₁₀. We have:
• n^th term = a + (n-1)d ⇒ a₁₀ = 17 + (10-1)×-1= 17 - 9 = 8
The AP can be seen here in a tabular form

Solved example 26.34
A manufacturer of TV sets produced 600 sets in the third year and 700 sets in the seventh year. Assuming that the production increases uniformly by a fixed number every year, find : (i) the production in the 1^st year (ii) the production in the 10^th year (iii) the total production in first 7 years
Solution:
• The production increases uniformly by a fixed number every year. So it is an AP
• We are given: a₃ = 600 and a₇ = 700
Part (i): We have to find the first term a
1. n^th term = a + (n-1)d ⇒ a₃ = 600 = a + (3-1)d ⇒ 600 = a + 2d
2. Again, n^th term = a + (n-1)d ⇒ a₇ = 700 = a + (7-1)d ⇒ 700 = a + 6d 
3. Now we have to solve the two equations.
(i) From (1) we have: a = 600-2d
(ii) Substituting this value of a in (2) we get:
700 = 600 - 2d + 6d ⇒ 700 = 600 + 4d ⇒ 4d = 100 ⇒ d = 25
(iii) substituting this value of d in (1) we get:
600 = a + 2 × 25 ⇒ 600 = a + 50 ⇒ a = 550
Part (ii): We have to find a₁₀
1. n^th term = a + (n-1)d ⇒ a₁₀ = 550 + (10-1)×25= 550 + 9 × 25 = 550 + 225 = 775
Part (iii): We have to find S₇
1. We have: S = ⁿ⁄₂ [2a + (n-1)d] = ⁷⁄₂ × [2 × 550 + (7-1)×25] 
= ⁷⁄₂ × [1100 + 6×25] = ⁷⁄₂ × [1100 +150] = ⁷⁄₂ × 1250 = 7 × 625 = 4375

Solved example 26.35
A contract on construction job specifies a 'penalty for delay of completion' beyond a certain date as follows: Rs 200 for the first day, Rs 250 for the second day, Rs 300 for the third day, etc., the penalty for each succeeding day being Rs 50 more than for the preceding day. How much money the contractor has to pay as penalty, if he has delayed the work by 30 days?
Solution:
Problem analysis:
■ The contractor has to complete a construction job on a specified date. Let this date be March 31^st.
• If the contractor completes the work only on April 1^st, it means that he has delayed the work by 1 day
    ♦ For this delay of 1 day, he has to pay Rs 200 
• If the contractor completes the work only on April 2^nd, it means that he has delayed the work by 2 days
    ♦ For this delay of  2 days, he has to pay Rs 200 + 250
• If the contractor completes the work only on April 3^rd, it means that he has delayed the work by 3 days
    ♦ For this delay of 3 days, he has to pay Rs 200 + 250 + 300
So on . . .
■ Thus we find that, it is an AP with a = 200 and d = 50
1. For a delay of 30 days, there will be 30 terms. So n = 30
2. We have: S = ⁿ⁄₂ [2a + (n-1)d] = ³⁰⁄₂ × [2 × 200 + (30-1)×50] 
= 15 × [400 + 29×50] = 15 × [400 +1450] = 15 × 1850 = Rs 27750

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In the next section we will see Sequences in general.

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Chapter 26.4 - Sum of first n terms of an Arithmetic progression

In the previous section we completed the discussion on how to determine the n^th term of an AP. We also saw some solved examples. In this section we will see how to determine the sum of first n terms of an AP.
Consider the following situation:
■ The mother of a child deposits a certain amount in a money box on every birthday of the child. 
• On the 1^st birthday she deposited Rs 100
• On the 2^nd birthday she deposited Rs 150
• On the 3^rd birthday she deposited Rs 200
• On the 4^th birthday she deposited Rs 250
So on . . .
• She continued to deposit in this way until the 21^st birthday. How much money will be there in the money box on the 21^st birthday?
Solution:
1. First of all we write the various deposits in the form of a series. We get:
100, 150, 200, 250, . . . 
2. We find that it is an AP. Because, every term can be obtained by adding a fixed number 50 to the preceding term.
3. So we have an AP with the 1^st term a = 100 and common difference d = 50
4. Now we can find the 21^st term
a₂₁ = 100 + (21 -1) × 50 = 100 + 20 × 50 = 100 + 1000 = 1100
5. Thus we have three items now:
• First term of the AP = a = 100
• Last term of the AP = a₂₁ = 1100
• Common difference of the AP = d = 50
6. But knowing these three items do not solve our problem. We want to know the total money in the box. For that we need to add all the terms of the AP. This is shown below:
Total money = 100 + 150 + 200 + . . . . + 1100
7. But this method is tedious and time consuming. Because, we have to write all the terms first and then add them one by one. There must be an easier way. 

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Let us see the method used by the famous mathematician Gauss to solve a problem which was given to him when he was just 10 years old:

■ The problem was this: Find the sum of the numbers from 1 to 100
• To solve it we have to write the step as:
1 + 2 + 3 + . . . + 100.
• We have to add the numbers one by one. But Gauss immediately gave the answer as 5050. How did he find the answer so soon?
■ His method is as follows: 
1. Let the required sum be 'S'. So first write: 
S = 1 + 2 + 3 + 4 + . . . + 100.
2. Just below that, write it in the reverse order like this:
S = 100 + 99 + 98 + . . . + 3 + 2 + 1
This is shown in the fig.26.2 below:

Fig.26.2

3. Both sums will be the same because the second is just the reverse of the first. So both are denoted as ‘S'
4. Now add the two rows together as shown in the fig.26.3 below:

Fig.26.3

We get 2S = 101 + 101 + 101 + . . . + 101
5. This series has 100 terms. Each term is 101. So the sum is 100 ×101 = 10100
6. So we can write: 2S = 10100. From this we get S = ¹⁰¹⁰⁰⁄₂ = 5050

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We will use a similar technique to find the sum of the first n terms of an AP. 

1. First we write it in the straight order:

S = a + (a+d) + (a+2d) + (a+3d) + . . . + [a+(n-3)d] + [a+(n-2)d] + [a+(n-1)d] 

2. Then, below it, in the reverse order:

[a+(n-1)d] + [a+(n-2)d] + [a+(n-3)d] + . . . + (a+3d) + (a+2d) + (a+d) + a

• The fig.26.4 below shows them in alignment: 

Fig.26.4

3. Next we add them. This is shown in fig.26.5 below:

Fig.26.5

4. The above addition is giving us interesting results. Let us analyse:
(i) Add the 1^st term of the 'straight series' with the 1^st term of the 'reversed series'. We get:
a + [a+(n-1)d] = 2a+(n-1)d
(ii) Add the 2^nd term of the 'straight series' with the 2^nd term of the 'reversed series'. We get:
(a+d) + [a+(n-2)d] = 2a+d+(n-2)d = 2a+d+nd-2d = 2a+nd-d = 2a+(n-1)d
• This is the same result in 4(i)
(iii) Add the 3^rd term of the 'straight series' with the 3^rd term of the 'reversed series'. We get:
(a+2d) + [a+(n-3)d] = 2a+2d+(n-3)d = 2a+2d+nd-3d = 2a+nd-d = 2a+(n-1)d
• This is the same result in 4(i)
So on . . . 
(iv) Add the last term of the 'straight series' with the last term of the 'reversed series'. We get:
[a+(n-1)d] + a = 2a+(n-1)d
• This is the same result in 4(i)
■ We will get the same result '2a+(n-1)d' for all the additions 
5. So we can write:
2S = n[2a+(n-1)d]
From this we get the equation for the sum of first n terms of an AP:
Eq.26.2:

6. The equation above can be written in another form:
• S = ⁿ⁄₂[2a+(n-1)d] same as S = ⁿ⁄₂[a + a +(n-1)d] 
• But a +(n-1)d is the n^th term a_n. So we get another form of the above equation:
Eq.26.2(a):

• That means, to find the sum of first n terms of an AP, we can follow a simple procedure:
(i) Add the first term and n^th term together
(ii) Multiply the sum by ⁿ⁄₂. This will be the required result
7. Yet another form:
• If there are only n terms in the given AP, the 'n^th term' is the last term l
• So eq.26.2(a) becomes:
Eq.26.2(b):

• That means, if there are only n terms in an AP, to find the sum of those n terms, we can follow a simple procedure:
(i) Add the first term and the last term l together
(ii) Multiply the sum by ⁿ⁄₂. This will be the required result
• This equation is useful if we are given the first term, last term and number of terms only
8. Another useful result:
• Let the sum of first n terms be S_n
• Let the sum of the first (n-1) terms be S_n-1
• Then the n^th term = S_n - S_n-1

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Now we can find the total money in the box. Let us use all the three equations and see if they give the same result.

1. Using eq.26.2:
S = ⁿ⁄₂ [2a + (n-1)d] ⇒ S = ²¹⁄₂ × [2 × 100 + (21-1)×50] 
= ²¹⁄₂ × [200 + 1000] = ²¹⁄₂ × [1200] = 21 × 600 = 12600

2. Using eq. 26.2(a):
S = ⁿ⁄₂ [a + a_n] ⇒ S = ²¹⁄₂ × [100 +  1100] = ²¹⁄₂ × 1200 = 21×600 = 12600

3. Using eq.26.2(b):
S = ⁿ⁄₂ [a + l]
This will obviously give the same result because the last term l = a_n = 1100, which is used in eq.26.2(a)

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So the equations involve 4 items: 

• The sum S, • the number of terms n,  • the first term a and • the common difference d.

If any three of these items are known, the fourth item can be easily calculated. Let us see some solved examples:

Solved example 26.26
Find the sum of the first 22 terms of the AP : 8, 3, –2, . . .
Solution:
1. In the given AP, a = 8, d = -2 -3 = -5 and n = 22
2. We have: S = ⁿ⁄₂ [2a + (n-1)d] ⇒ S = ²²⁄₂ × [2 × 8 + (22-1)×(-5)]
= ²²⁄₂ × [16 - 105] = ²²⁄₂ × [-89] = 11 × -89 = -979

Solved example 26.27
If the sum of the first 14 terms of an AP is 1050 and its first term is 10, find the 20^th term.
Solution:
1. Given: S₁₄ = 1050, a = 10
2. We have: S = ⁿ⁄₂ [2a + (n-1)d] ⇒ S₁₄ = ¹⁴⁄₂ × [2 × 10 + (14-1)×d]
⇒ 1050 = 7 × [20+13d] ⇒ 1050 = 140 + 91d ⇒ 91d = 910 ⇒ d = 10 
3. Now we have a and d. Using those, we can calculate any a_n
• We have: a_n = a + (n-1)d
• So a₂₀ = 10 + 19 × 10 = 10 + 190 = 200

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In the next section we will see a few more solved examples.

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Chapter 26.3 - Arithmetic progressions - Solved examples

In the previous section we learned how to determine the n^th term of an AP. We also saw some solved examples. In this section we will see a few more solved examples.

Solved example 26.15
Find the number of terms in the following APs:
(i) 7, 13, 19, . . . , 205
Solution:
1. This is an AP with a = 7 and d = 6. Because every term can be obtained by adding 6 to the preceding term.
3. Let n be the total number of terms. We have to find the n value of the last term 205
4. We have:  n^th term = a + (n-1)d ⇒ 205 = 7 + (n-1) ×6 
⇒  205 = 7 + 6n - 6 ⇒ 6n = 205 - 7 + 6  ⇒ 6n = 204 ⇒ n = 34
• So there are a total of 34 terms in the given AP.
(ii) 18, 15¹⁄₂, 13, . . . , -47
Solution:
1. This is an AP with a = 18 and d = -2¹⁄₂. = -⁵⁄₂. Because, every term can be obtained by adding (-2¹⁄₂) to the preceding term.
3. Let n be the total number of terms. We have to find the n value of the last term -47
4. We have:  n^th term = a + (n-1)d ⇒ -47  = 18 + (n-1) ×-⁵⁄₂ 
⇒  -47 = 18 -⁵ⁿ⁄₂ + ⁵⁄₂ ⇒ ⁵ⁿ⁄₂ = 47 + 18 +⁵⁄₂  ⇒ 5n = 94 + 36 + 5 ⇒ 5n = 135 ⇒ n = 27

• So there are a total of 27 terms in the given AP.

Solved example 26.16
Check whether -150 is a term in the AP: 11, 8, 5, 2, . . . 
Solution:
1. 1^st term a = 11
2. Common difference d = 2 - 5 = -3
3. We have: n^th term = a + (n-1)d
4. Let -150 be the n^th term. Then substituting the values in (3) we get:
5. -150 = 11 + (n-1) × -3 ⇒ -150 = 11 - 3n + 3 ⇒ 3n = 150 + 11 +3 ⇒ 3n = 164 ⇒ n = ¹⁶⁴⁄₃
• 'n' is a number. It shows a position. Like 1^st, 2^nd, 3^rd etc., So 'n' should be a whole number.

• ¹⁶⁴⁄₃ is not a whole number. So -150 is not a term in the given AP. 

Solved example 26.17
Determine the 31^st term of an AP whose 11^th term is 38 and the 16^th term is 73.   
Solution:
1. Let a be the 1^st term and d the common difference
Then we have: n^th term = a + (n-1)d
2. Substituting the values we get:
(i) 38 = a + (11-1) × d ⇒ 38 = a + 10d
(ii) 73 = a + (16-1) × d ⇒ 73 = a + 15d    
3. So we have two equations in two variables a and d. We can easily solve them as follows (Details here):
4. From (i) we get a = 38 - 10d
5. Substituting this in (ii) we get: 73 = (38 - 10d) + 15d 
⇒ 73 = 38 + 5d ⇒ 5d = 35 ⇒ d = 7
6. Substituting this value of d in (4) we get: a = 38 - 10 × 7  ⇒ a = 38 -70 ⇒ a = -32
7. So we have a and d. We can now find any term of the AP.
• we have: n^th term = a + (n-1)d
• So a₃₁ = -32 + (31-1) ×7 = -32 + 30 × 7 = -32 + 210 = 178

Solved example 26.18
If the 3^rd and the 9^th terms of an AP are 4 and -8 respectively, which term of this AP is zero?
Solution:
1. Let a be the 1^st term and d the common difference
Then we have: n^th term = a + (n-1)d
2. Substituting the values we get:
(i) 4 = a + (3-1) × d ⇒ 4 = a + 2d
(ii) -8 = a + (9-1) × d ⇒ -8 = a + 8d    
3. So we have two equations in two variables a and d. We can easily solve them as follows (Details here):
4. From (i) we get a = 4 - 2d
5. Substituting this in (ii) we get: -8 = (4 - 2d) + 8d 
⇒ -8 = 4 + 6d ⇒ 6d = -12 ⇒ d = -2
6. Substituting this value of d in (4) we get: a = 4 - 2 × (-2)  ⇒ a = 4 + 4 ⇒ a = 8
7. So we have a and d. We can now find any term of the AP.
• we have: n^th term = a + (n-1)d

• So 0 = 8 + (n-1) ×-2 ⇒ 0 = 8 -2n + 2 ⇒ 2n = 10 ⇒ n = 5
• Thus zero is the 5^th term of the AP

Solved example 26.19
The 17^th term of an AP exceeds its 10^th term by 7. Find the common difference
Solution:
1. Let a be the 1^st term and d the common difference
Then we have: n^th term = a + (n-1)d
2. Substituting the values we get:
(i) a₁₇ = a + (17-1) × d ⇒ a₁₇ = a + 16d
(ii) a₁₀ = a + (10-1) × d ⇒ a₁₀ = a + 9d    
3. Given that: a₁₀ + 7 = a₁₇. Substituting this in (2) we get:
4. a + 9d + 7 = a + 16d ⇒ 9d + 7 = 16d ⇒ 7d = 7 ⇒ d = 1

Solved example 26.20
Which term of the AP : 3, 15, 27, 39, . . . will be 132 more than its 54^th term?
Solution:
1. Let a be the 1^st term and d the common difference
Then we have: n^th term = a + (n-1)d
2. So 54^th term = a₅₄ = a + (54-1) × d ⇒ a₅₄ = a + 53d 
3. Let a_n be the required term. We have: a_n = a + (n-1)d  
4. According to the given condition, a_n = a₅₄ + 132
5. Substituting (2) and (3) in (4) we get:
a + (n-1)d = a + 53d + 132 ⇒ (n-1)d = 53d + 132 
⇒ nd - d = 53d + 132 ⇒ (n-1-53)d = 132 ⇒ (n-54)d = 132
6. But from the given AP, d = 39-27 = 12
7. So from (5) we get (n-54) × 12 = 132 ⇒ n-54 = ¹³²⁄₁₂ = 11 ⇒ n = 11 + 54 = 65

Solved example 26.21
Two APs have the same common difference. The difference between 100^th terms is 100, what is the difference between their 1000^th terms?
Solution:
1. The two APs have the same common difference. Let it be d
2. Let the 1^st terms be a₁ and b₁ respectively
3. Then for the first AP: a₁₀₀ = a₁ + 99d 
4. For the second AP: b₁₀₀ = b₁ + 99d  
5. Given that the difference of the 100^th term is 100. So we can write:
(a₁ + 99d) - (b₁ + 99d) = 100 ⇒ a₁ + 99d - b₁ - 99d = 100 ⇒ a₁ - b₁ = 100
6. Now we write the 1000^th terms:
7. For the first AP: a₁₀₀₀ = a₁ + 999d 
8. For the second AP: b₁₀₀₀ = b₁ + 999d
9. Difference = (a₁ + 999d) - (b₁ + 999d) = a₁ - b₁.
10. But from (5) we have: a₁ - b₁ = 100
11. So the difference between their 1000^th terms = 100

Solved example 26.22
How many three-digit numbers are divisible by 7?
Solution:
• There are so many 3 digit numbers. The list is very long but not endless. All the 3 digit numbers fall between 100 and 999 both of them inclusive. 
• Out of them, many are divisible by 7. In this problem, we do not have to actually pick out all those which are divisible by 7. 
• All we need is the 'number 'of 3-digit numbers which are divisible by 7. Let us try to find it:
1. The first 3-digit number is 100. It is not divisible by 7
• If we divide 100 by 7, we get a reminder 2. [100 = (14 × 7) + 2 = 98 + 2]
• So the '2' is causing the problem. If this '2' was not there, the number would be divisible by 7
• To avoid '2' we can take (100-2) = 98. It is divisible by 7. 
• But 98 is a 2-digit number. It falls outside our required range  
• Then how do we avoid the problem causing '2'?
• Let us make this '2' into a '7'.
• When a number is divided by '7', if the remainder is '7', then it means that there is no remainder,
• So to get '7' as remainder instead of '2', we must add (7-2) = 5 to the original number
• Thus the next number above 100 which is divisible by 7 is 100 + (7-2) = 100 + 5 = 105
2. The last 3-digit number is 999. It is not divisible by 7
• If we divide 999 by 7, we get a reminder 5. [999 = (142 × 7) + 5 = 994 + 5]
• So the '5' is causing the problem. If this '5' was not there, the number would be divisible by 7

• To avoid '5' we can take (999-5) = 994. It is divisible by 7. 
Note: We could make remainder '5' into remainder '7' by adding (7-5) = 2 to 999
That is we take (2+999) = 1001. It is divisible by 7. (1001 ÷ 7 = 143) 
But 1001 is a 4-digit number. It falls outside our required range 
• So the number below 999 which is divisible by 7 is 999 - 5 = 994
The above results can be seen in a tabular form here.
3. Thus the 1^st term of the AP is 105 and the last term is 994. The common difference d is 7
4. So we can write:  n^th term = a + (n-1)d ⇒  n^th term = 105 + (n-1) ×7   
6. Let 994 be the n^th term. Then we can write: 
994 =  105 + (n-1) ×7 ⇒ 994 = 105 + 7n -7 ⇒ 7n = 994 - 98 ⇒ 7n = 896 ⇒ n = 128
7. So the last term 994 is the 128^th term. We can write:
There are 128 three-digit numbers which are divisible by 7

Solved example 26.23
How many multiples of 4 lie between 10 and 250?
Solution:
1. The first number given is 10. It is not divisible by 4
• If we divide 10 by 4, we get a reminder 2. [10 = (4 × 2) + 2 = 8 + 2] 
• So the next number above 10 which is divisible by 2 is 10 + (4-2) = 10 + 2 = 12
2. The last number given is 250. It is not divisible by 4
• If we divide 250 by 4, we get a reminder 2. [250 = (4 × 62) + 2 = 248 + 2] 
• So the number below 250 which is divisible by 2 is 250 - 2 = 248
3. Thus the 1^st term of the AP is 12 and the last term is 248. The common difference d is 4
4. So we can write:  n^th term = a + (n-1)d ⇒  n^th term = 12 + (n-1) ×4   
6. Let 248 be the n^th term. Then we can write: 
248 =  12 + (n-1) ×4 ⇒ 248 = 12 + 4n -4 ⇒ 4n = 248 - 8 ⇒ 4n = 240 ⇒ n = 60
7. So the last term 248 is the 60^th term. We can write:
• There are 60 numbers between 10 and 250,which are divisible by 4. This is same as:
• There are 60 multiples of 4 between 10 and 250

Solved example 26.24      

For what value of n, are the nth terms of two APs: 63, 65, 67, . . . and 3, 10, 17, . . . equal?
Solution:
1. We can write:
• n^th term of the first AP = 63 + (n-1) × 2 = 61 + 2n [∵ d = 67-65 = 2]
• n^th term of the second AP = 3 + (n-1) × 7 = 7n - 4 [∵ d = 17-10 = 7]
2. If these two terms are equal, we can write: 
61 + 2n = 7n -4 ⇒ 5n = 65 ⇒ n = 13

Solved example 26.25
The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of the AP.
Solution:
1. 4^th term = a + 3d, 8^th term = a + 7d
Sum = 2a + 10d = 24 ⇒ a + 5d = 12
2. 6^th term = a + 5d, 10^th term = a + 9d
Sum = 2a + 14d = 44 ⇒ a + 7d = 22
3. From (1) we get: a = 12-5d. Substituting this in (2) we get:
12 - 5d + 7d = 22 ⇒ 2d = 10 ⇒ d = 5
4. Substituting this value of d in (1) we get:
a + 5 × 5 = 12 ⇒ a + 25 = 12 ⇒ a = -13
5. So we have a and d. We can now form the AP.
1^st term = a₁ = a = -13  
2^nd term = a₂ = a + d  = -13 + 5 = -8
3^rd term = a₃ = a + 2d  = -13 + 2×5 =  -13 + 10 = -3
Thus the AP is: -13, -8, -3, . . .

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In the next section we will sum of first n terms of an AP.

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