In the previous section we completed the discussion on how to determine the nth 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 1st birthday she deposited Rs 100
• On the 2nd birthday she deposited Rs 150
• On the 3rd birthday she deposited Rs 200
• On the 4th birthday she deposited Rs 250
So on . . .
• She continued to deposit in this way until the 21st birthday. How much money will be there in the money box on the 21st 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 1st term a = 100 and common difference d = 50
4. Now we can find the 21st 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.
• 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:
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:
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
3. Next we add them. This is shown in fig.26.5 below:
4. The above addition is giving us interesting results. Let us analyse:
(i) Add the 1st term of the 'straight series' with the 1st term of the 'reversed series'. We get:
a + [a+(n-1)d] = 2a+(n-1)d
(ii) Add the 2nd term of the 'straight series' with the 2nd 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 3rd term of the 'straight series' with the 3rd 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 nth term an. 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 nth 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 'nth 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 Sn
• Let the sum of the first (n-1) terms be Sn-1
• Then the nth term = Sn - Sn-1
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 + an] ⇒ 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 = an = 1100, which is used in eq.26.2(a)
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 20th 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 an
• We have: an = a + (n-1)d
• So a20 = 10 + 19 × 10 = 10 + 190 = 200
In the next section we will see a few more solved examples.
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 1st birthday she deposited Rs 100
• On the 2nd birthday she deposited Rs 150
• On the 3rd birthday she deposited Rs 200
• On the 4th birthday she deposited Rs 250
So on . . .
• She continued to deposit in this way until the 21st birthday. How much money will be there in the money box on the 21st 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 1st term a = 100 and common difference d = 50
4. Now we can find the 21st 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.
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 |
4. Now add the two rows together as shown in the fig.26.3 below:
 |
| Fig.26.3 |
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
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 |
 |
| Fig.26.5 |
(i) Add the 1st term of the 'straight series' with the 1st term of the 'reversed series'. We get:
a + [a+(n-1)d] = 2a+(n-1)d
(ii) Add the 2nd term of the 'straight series' with the 2nd 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 3rd term of the 'straight series' with the 3rd 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:
• S = n⁄2[2a+(n-1)d] same as S = n⁄2[a + a +(n-1)d]
• But a +(n-1)d is the nth term an. So we get another form of the above equation:
Eq.26.2(a):
(i) Add the first term and nth 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 'nth term' is the last term l
• So eq.26.2(a) becomes:
Eq.26.2(b):
(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 Sn
• Let the sum of the first (n-1) terms be Sn-1
• Then the nth term = Sn - Sn-1
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 + an] ⇒ 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 = an = 1100, which is used in eq.26.2(a)
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 20th 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 an
• We have: an = a + (n-1)d
• So a20 = 10 + 19 × 10 = 10 + 190 = 200
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