Simple explanation and solution to the arithmetic progression
What is an arithmetic progression?
Observe these series:
Ex(1): 1, 2, 3, 4, 5, 6...
Ex(2): 2, 4, 6, 8, 10 ...
Ex(3): 1/2, 3/2, 5/2, 7/2...
In example 1, you can see the difference between two consecutive terms: 2-1 = 1, 3-2 = 1, 4-3 = 1... and so on.
In example 2: 4-2 = 2
6-4 = 2
8-6 = 2
and so on. Similarly, the difference of the two terms in example 3 is 1/2.
A series in which consecutive terms differ by a constant (fixed) value is called arithmetic series.
The first term is often represented by T1 or a, and the fixed difference is denoted by "d"
If we observe this type of series, we can write it as:
a, (a+d), (a+2d), (a+3d),...
The first term contains no d, or it may be written as (1-1)d
The second term contains one d and can be written as (2-1)d
The third term contains 2d and can be written as (3-1)d.
....
In this manner, we can write the formula for the 10th term as a + (10-1)d.
So, in general, the nth term from the beginning can be written
Tn = a + (n-1)d
If we write n = 1, 2, 3, 4...
We get the values of the first terms, T1, T2, T3, T4, and so on.
If we reverse the series and write from last to first, then:
Tn' = Tn +(n-1)(-d)
The negative sign is present because we have reversed the series.
Tn' is the last term on reversing the series.
Example: 3, 5, 7, 9, and 11 is a given series. Common difference = 2
Reversing the series, we get 11, 9, 7, 5, 3; common difference (d) = -2
Solutions
[2]-[1]
2d = -10
d = -5
putting the value of d in [1]
a = 3-d
= 3-(-5)
=8
Fourth term
T4 = a + 3d
=8 +3*(-5)
= 8-15
=-7
(iii) 5, -, -, 9 1/2
Here, the number of terms is 4.
a = 5, d?, T4 = 9 1/2
T4 = a + 3d
19/2 = 5 + 3d
19/2 - 5 = 3d
9/2 = 3d
d = 3/2
T2 = a+d = 5 + 3/2 = 13/2
T3 = a + 2d = 5 + 2 * 3/2
= 8
(iv)-4, -, -, -, -, 6
Total number of terms = 6
a = -4
T6= 6
6 = -4 + 5d
6+4 = 5d
d = 10/5 = 2
T2 = a+d = -4+2 = -2
T3 = T2 + d = -2 + 2 = 0
T4 = T3 + d = 0 + 2 = 2
T5 = 2+2 = 4
(v) T2 = 38
T6 = 22
Writing an equation for these conditions
T2: a+d = 38 ...[1]
T6: a+5d = 22...[2]
[2]-[1]
4d = 22-38 = -16
d = -16/4 = -4
from [1]
a+(-4) = 38
a = 38 + 4 = 42
T3 = T2 + d = 38 + (-4) = 34
T4 = T3+d = 34+(-4) = 30
T5 = T4 + d = 30 + (-4) = 26
Question 4
Let Tn = 78
a = 3
d = 8-3 = 5
78 = 3 + (n-1)*5
78-3 = (n-1)*5
75/5 = (n-1)
15 = n-1
n = 15 + 1 = 16
Question 5
(i) 7, 13, 19, ..., 205
a = 7, d = 13-7 = 6
Let Tn = 205
205 = 7 + (n-1)*6
205-7 = (n-1)6
198/6 = n-1
33 = n-1
n=34
34th term is 205
(ii) 18, 15 1/2, ... -47
a = 18
d = 31/2 - 18
= -5/2
Tn = a + (n-1)d
-47 = 18 + (n-1)(-5/2)
-65 = (n-1)(-5/2)
-65*(-2/5) = n-1
26 + 1 = n
n = 27
27th term is -47
Q6. 11, 8, 5, -2,...
Let Tn = -150.
a = 11
d = 8-1 =-3
-150 = 11 + (n-1)(-3)
-161 = (n-1)(-3)
(n-1) = -161/-3 = not an integer; hence, it is not a term of this A.P.
7. Given T11 = 38 and T16 =73, find T38
Let the first term be a, the common difference be d
T11 =
a + 10d = 38 (1)
a + 15d = 73 (2)
Subtract (1) from (2)
5d = 35
d = 35/5 = 7
from (1) a + 7 * 10 = 38
a = -32
T38 = a + (38-1)d
=-32 + 37*7
=-32 + 259
= 229
Q8
Q9: Given T3 = 4
And
T9 = -8
Then,
a +2d =4 … .(1)
T9:
a+8d = -8 …(2)
(2)
–(1)
6d
=-12
ð
d
=-2, from (1) a+(-4) = 4
ð
a=8
ð
let
Tn =0
ð
Then,8
+(n-1)*(-2) = 0
ð
8-2n+2
= 0
ð
10
=2n
ð
10/2
= n, fifth term is zero
Q10
Seventeenth
term = tenth term +7
A
+16d = A+9d +7
ð
7d
=7
ð
D=
1
ð
Common
difference is 1
Q11: a =3, d = 15-3 =12
Let
Tn = t54 +132
A+(n-1)d
=A+53d +132
(n-1)*12
=53*12 +132
(n-1)*12
= 636+132
(n-1)*12
= 768
(n-1)
= 768/12 =64
Therefore,
n = 64+1 = 65
65th
term
Q12:
Let the first term of the first A.P. be x, and the first term of the second A.P.
be y, and the common difference be d
100th
term of first A.P. -100th
term of the 2nd AP= 100
x+99d
–(y+99d) =100
x-y
=100
Since
the difference of the 100th term depends on the difference of their
first term, so the difference of the 1000th term will be the same = 100
Q13:
First three digit number divisible by 7 = 7*15 = 105
Common
difference =7
The largest three-digit number divisible can be obtained by the trial method
140*7
= 980, 141*7 = 987, 142*7 = 994
Let
the nth term be 994
Then,
994 =105 +(n-1)*7
994-105
=7n-7
889+7
= 7n
896
=7n
=
896/7 = 128
Hence,
there are 128 terms
Q14:
First two digit multiple of 4 =12, largest 2 digit number below 250 = 248,
common difference =4
therefore, 248 = 12 +(n-1)*4
248-12=
4n-4
236
+4 = 4n
240/4
= n
60
=n
Number
of terms divisible by 4 =60
Q15: nth term of 63, 65 67 …= 63 +(n-1)2 { here, d= 65-63 =2}
Nth term of the 2nd sequence= 3 +(n-1)7 { Common difference = 10-3 =7}
According to the question
63+(n-1)2
= 3 +(n-1)7
63
+ 2n-2 = 3 +7n -7
61-3+7
= 7n-2n
65
= 5n
65/5
=n
13
=n
16. Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12.
Solution:
Let the first term be a and the common difference be d.
Third term: a + 2d = 16. (1)
Difference between 7th and 5th terms: (a+6d) - (a+4d) = 2d = 12.
So d = 6. Substitute in (1): a + 2(6) = 16 ⇒ a + 12 = 16 ⇒ a = 4.
Therefore, the AP is: 4, 10, 16, 22, ... (first term 4, common difference 6).
17. Find the 20th term from the last term of the AP: 3, 8, 13, …, 253.
Solution:
Here a₁ = 3, d = 5, last term l = 253.
Number of terms N satisfies l = a₁ + (N-1)d ⇒ 253 = 3 + (N-1)·5.
So 250 = 5(N-1) ⇒ N-1 = 50 ⇒ N = 51.
The 20th term from the last is the term numbered N - 19 (since the last is the 1st from last). So index = 51 - 19 = 32.
32nd term: a₃₂ = a₁ + (32-1)d = 3 + 31·5 = 3 + 155 = 158.
Answer: 158.
18. 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:
Let the first term be a and the common difference be d.
4th + 8th: (a+3d) + (a+7d) = 2a + 10d = 24 ⇒ a + 5d = 12. (A)
6th + 10th: (a+5d) + (a+9d) = 2a + 14d = 44 ⇒ a + 7d = 22. (B)
Subtract (A) from (B): (a+7d) - (a+5d) = 2d = 10 ⇒ d = 5.
From (A): a + 5·5 = 12 ⇒ a + 25 = 12 ⇒ a = -13.
The first three terms: a, a+d, a+2d = -13, -8, -3.
Answer: -13, -8, -3.
19. Subba Rao started work in 1995 at an annual salary of ₹5000 and received an increment of ₹200 each year. In which year did his income reach ₹7000?
Solution:
Salary after n increments: 5000 + 200n. In 1995, he had n = 0.
Set 5000 + 200n = 7000 ⇒ 200n = 2000 ⇒ n = 10.
Since n = 0 corresponds to 1995, the year when n = 10 is 1995 + 10 = 2005.
Answer: His income reached ₹7000 in the year 2005.
20. Ramkali saved ₹5 in the first week of the year and then increased her weekly savings by ₹1.75. If in the nth week her weekly savings become ₹20.75, find n.
Solution:
This is an AP with a = 5 and d = 1.75. We want a + (n-1)d = 20.75.
So 5 + (n-1)·1.75 = 20.75 ⇒ (n-1)·1.75 = 15.75 ⇒ n-1 = 15.75 / 1.75 = 9.
Thus n = 10.
Answer: She saved ₹20.75 in the 10th week.
Exercise 5.3 — Solutions (Arithmetic Progressions)
1. Find the sum of the following APs.
(i) 2, 7, 12, … (10 terms)
Work: a = 2, d = 5, n = 10. Use S_n = n/2 [2a + (n−1)d].
S_{10} = 10/2 [2·2 + 9·5] = 5 [4 + 45] = 5·49 = 245.
(ii) −37, −33, −29, … (12 terms)
Work: a = −37, d = 4, n = 12.
S_{12} = 12/2 [2(−37) + 11·4] = 6 [−74 + 44] = 6·(−30) = −180.
(iii) 0.6, 1.7, 2.8, … (100 terms)
Work: a = 0.6, d = 1.1, n = 100.
S_{100} = 100/2 [2·0.6 + 99·1.1] = 50 [1.2 + 108.9] = 50·110.1 = 5505.
(iv) 1/15, 1/12, 1/10, … (11 terms)
Work: a = 1/15, next term 1/12 so d = 1/12 − 1/15 = 1/60, n = 11.
S_{11} = 11/2 [2·(1/15) + 10·(1/60)] = 11/2 [2/15 + 10/60] = 11/2 [2/15 + 1/6].
Simplify: 2/15 + 1/6 = 4/30 + 5/30 = 9/30 = 3/10, so S_{11} = 11/2 · 3/10 = 33/20 = 1.65.
2. Find the sums given below
(ii) 34 + 32 + 30 + … + 10
Work: a = 34, l = 10, d = −2. Number of terms n = (l−a)/d + 1 = (10−34)/(−2)+1 = 13.
S = n(a + l)/2 = 13(34 + 10)/2 = 13·44/2 = 13·22 = 286.
(iii) −5 + (−8) + (−11) + … + (−230)
Work: a = −5, d = −3, last l = −230. Number of terms n = (l−a)/d +1 = (−230 +5)/(−3) +1 = 76.
S = n(a + l)/2 = 76(−5 −230)/2 = 76(−235)/2 = 38(−235) = −8930.
3. In an AP … (multiple small parts)
(i) Given a = 5, d = 3, a_n = 50. Find n and S_n.
Work: a_n = a + (n−1)d ⇒ n = ((50−5)/3) + 1 = 16.
S_{16} = 16/2 [2·5 + 15·3] = 8[10 + 45] = 8·55 = 440.
(ii) Given a = 7, a_{13} = 35. Find d and S_{13}.
Work: d = (35 − 7)/12 = 28/12 = 7/3.
S_{13} = 13/2 (a + a_{13}) = 13/2 (7 + 35) = 13/2 · 42 = 13·21 = 273.
(iii) Given a_{12} = 37, d = 3. Find a and S_{12}.
Work: a = a_{12} − 11d = 37 − 33 = 4.
S_{12} = 12/2 [2·4 + 11·3] = 6 [8 + 33] = 6·41 = 246.
(iv) Given a_3 = 15, S_{10} = 125. Find d and a_{10}.
Work: a + 2d = 15 ⇒ a = 15 − 2d. Use S10 = 10/2 [2a + 9d] = 125 ⇒ 5(2a + 9d) = 125 ⇒ 2a + 9d = 25.
Substitute a: 2(15 − 2d) + 9d = 25 ⇒ 30 − 4d + 9d = 25 ⇒ 5d = −5 ⇒ d = −1.
Then a = 15 − 2(−1) = 17, and a_{10} = a + 9d = 17 − 9 = 8.
(v) Given d = 5, S_9 = 75. Find a and a_9.
Work: S9 = 9/2 [2a + 8d] = 75 ⇒ (9/2)(2a + 40) = 75 ⇒ 2a + 40 = 150/9 = 50/3.
So 2a = 50/3 − 40 = −70/3 ⇒ a = −35/3 ≈ −11.666…
Then a_9 = a + 8d = −35/3 + 40 = 85/3 ≈ 28.333…
(vi) Given a = 2, d = 8, S_n = 90. Find n and a_n.
Work: Solve n/2 [2·2 + (n−1)8] = 90. Solving gives n = 5.
Then a_5 = 2 + 4·8 = 34.
(vii) Given a = 8, a_n = 62, S_n = 210. Find n and d.
Work: Use S_n = n/2 (a + a_n) ⇒ 210 = n/2 (8 + 62) = n·35 ⇒ n = 6.
Then d = (62 − 8)/(n − 1) = 54/5 = 10.8.
(viii) Given a_n = 4, d = 2, S_n = −14. Find n and a.
Work: Use S_n = n/2 (a + a_n) = −14 and a = a_n − (n−1)d.
Solving gives n = 7 and a = −8.
(ix) Given a = 3, n = 8, S = 192. Find d.
Work: 192 = 8/2 [2·3 + 7d] ⇒ 192 = 4(6 + 7d) ⇒ 6 + 7d = 48 ⇒ d = 6.
(x) Given last term l = 28, S = 144, and number of terms = 9. Find a.
Work: S = n/2 (a + l) ⇒ 144 = 9/2 (a + 28) ⇒ a + 28 = 288/9 = 32 ⇒ a = 4.
4. How many terms of the AP 9, 17, 25, … must be taken to give a sum of 636?
Work: a = 9, d = 8. Solve n/2 [2·9 + (n−1)8] = 636.
Solving the quadratic gives n = 12 (positive integer solution).
5. The first term of an AP is 5, the last term is 45 and the sum is 400. Find n and d.
Work: S = n/2 (a + l)
⇒ 400 = n/2 (5 + 45) = n·25
⇒ n = 16.
d = (l − a)/(n − 1)
= 40/15
= 8/3
6. First term = 17, last term = 350, common difference = 9. How many terms and what is the sum?
Work: Number of terms n = (350 − 17)/9 + 1
= 38.
S = n/2 (a + l)
= 38/2 (17 + 350)
= 6973.
7. Find the sum of the first 22 terms of an AP in which d = 7 and the 22nd term is 149.
Work: a_{22} = a + 21d = 149
⇒ a = 149 − 21·7
= 2.
S{22} = 22/2 [2·2 + 21·7]
= 11 [4 + 147]
= 11·151
= 1661.
8. Sum of the first 51 terms where the second and third terms are 14 and 18, respectively.
Work: a + d = 14
and a + 2d = 18
⇒ d = 4,
a = 10.
S{51} = 51/2 [2·10 + 50·4]
= 51/2 [20 + 200]
= 51/2 · 220
= 51·110
= 5610.
9. If S7 = 49 and S17 = 289, find Sn (sum of first n terms).
Work: From S7 and S17 we get the system:
7/2(2a + 6d) = 49
⇒ 2a + 6d = 14
17/2(2a + 16d) = 289
⇒ 2a + 16d = 34
Subtract: 10d = 20 ⇒ d = 2. Then 2a + 12 = 14 ⇒ a = 1.
Therefore S_n = n/2 [2·1 + (n−1)·2] = n/2 · 2n = n^2. (So the sum of the first n terms is n².)
10. Show that the sequences are APs and find the sum of the first 15 terms.
(i) a_n = 3 + 4n
Work: Terms differ by 4 becausea_{n+1} − a_n = 4, so it's an AP with d = 4.
Sum of first 15: compute Σ (3 + 4n), S_{15} = 15/2 [2·7 + 14·4]
= 15/2 [14 + 56]
= 15/2 ·70
= 15*35
= 525.
(ii) a_n = 9 − 5n
Work: a_{n+1} − a_n = −5,
so AP with d = −5.
First term a_1 = 4.
S_{15} = 15/2 [2*4 + 14*(−5)]
= 15/2 [8 − 70]
= 15/2 ·(−62)
= 15*(−31) = −465.
11. If Sn = 4n − n², find first term, second term etc.
Work: S_1 = 4(1) − 1² = 3 so first term = 3.
S_2 = 4·2 − 2² = 8 − 4 = 4. So second term = t_2 = S_2 − S_1 = 4 − 3 = 1.
General term: t_n =
S_n − S_{n−1} = (4n − n²) − [4(n−1) − (n−1)²]
= 5 − 2n.
So t_3 = 5 − 2*3 = −1,
t_{10} = 5 − 20 = −15,
and the nth term is 5 − 2n.
12. Sum of the first 40 positive integers divisible by 6.
Work: These are 6, 12, …, 6·40 = 240.
Factor out 6: sum = 6(1 + 2 + … + 40) = 6·40·41/2 = 6·820 = 4920.
13. Sum of first 15 multiples of 8.
Work: 8(1 + … + 15) = 8·15·16/2 = 8·120 = 960.
14. Sum of odd numbers between 0 and 50.
Work: Odd numbers: 1, 3, 5, …, 49. There are 25 terms. Sum of first n odd numbers = n². Thus sum = 25² = 625.
15. Penalty for delay: Rs. 200 for 1st day, 250 for 2nd, 300 for 3rd, … (increase Rs. 50 each day). How much for 30 days?
Work: a = 200, d = 50, n = 30.
S_{30} = 30/2 [2·200 + 29·50] = 15[400 + 1450] = 15·1850 = 27750.
Contractor pays Rs. 27,750.
16. A sum of Rs 700 is to be given to seven prizes. Each prize is Rs 20 less than the previous one. Find each prize.
Work: Let the first prize be P, d = −20, n = 7, sum = 700.
700 = 7/2 [2P + 6(−20)]
⇒ (7/2)(2P − 120) = 700
⇒ 2P − 120 = 200
⇒ 2P = 320 ⇒ P = 160.
Prizes are: 160, 140, 120, 100, 80, 60, 40
(sum = 700).
17. Tree plantation: classes I to XII. Each section of class k plants k trees. Three sections per class. How many trees in total?
Work: For class k, each section plants k trees; there are 3 sections: class k contributes 3k trees. Sum for k = 1..12 is
class 1 plant = 3 , class 2 will plant = 2*3 =6 , ... a= 3, d =3, T12 =3*12 =36
S12 = 12/2(3+36) = 6*39 = 234
19. 200 logs stacked: bottom row 20, next 19, next 18, … . In how many rows are the 200 logs placed and how many in the top row?
Work: Rows form an AP: 20, 19, 18, … with difference −1. Let n be number of rows and top row has 20 − (n − 1) logs. Sum = n/2 [first + last] = 200.
Solving gives two algebraic roots but only integer valid is n = 16. Then top row = 20 − 15 = 5.
Answer: 16 rows; top row has 5 logs.
20. Potato race: bucket at start, first potato is 5 m from it, other potatoes are 3 m apart, ten potatoes in the line. Find positions.
Work: Distances from bucket form AP: 5, 8, 11, … with a = 5, d = 3, n = 10.
Positions (in metres):
5, 8, 11, 14, 17, 20, 23, 26, 29, 32.
Sn = 10/2(2*5 + (10-1)*3) = 5*(10+27) = 185m
For
n =13
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