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 Sn = 78
2. We have to find n
3. We have: Sn = 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: nth 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 23rd 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: nth 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 76th 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, an = 50, find n and Sn.
Solution:
1. In this problem, the nth term an is given as 50.
2. The sum from the first term 5 to the nth 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: nth 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 16th 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, an = 62, Sn = 210, find n and d.
Solution:
1. We know the first term, last term and Sn. 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 6th term is 62
2. We can use eq.26.1 which gives the position of any term:
• nth 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:
• nth 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:
• nth 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 1st year (ii) the production in the 10th 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. nth term = a + (n-1)d ⇒ a3 = 600 = a + (3-1)d ⇒ 600 = a + 2d
2. Again, nth 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. nth 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 31st.
• If the contractor completes the work only on April 1st, 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 2nd, 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 3rd, 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
In the next section we will see Sequences in general.
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 Sn = 78
2. We have to find n
3. We have: Sn = 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: nth 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 23rd 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: nth 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 76th 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, an = 50, find n and Sn.
Solution:
1. In this problem, the nth term an is given as 50.
2. The sum from the first term 5 to the nth 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: nth 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 16th 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, an = 62, Sn = 210, find n and d.
Solution:
1. We know the first term, last term and Sn. 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 6th term is 62
2. We can use eq.26.1 which gives the position of any term:
• nth 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:
• nth 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:
• nth 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 1st year (ii) the production in the 10th 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. nth term = a + (n-1)d ⇒ a3 = 600 = a + (3-1)d ⇒ 600 = a + 2d
2. Again, nth 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. nth 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 31st.
• If the contractor completes the work only on April 1st, 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 2nd, 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 3rd, 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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