Chapter 5 Class 10 Mathematics Arithmetic Progressions (AP)

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Chapter 5 in Class 10 Mathematics usually covers topics related to Arithmetic Progressions (AP). The chapter typically includes exercises on how to find the nth term of an arithmetic progression, sum of the first n terms of an AP, and word problems related to the same.

Let’s go through the steps involved in solving the exercises and solve the problems in detail.

Chapter 5: Arithmetic Progressions

Exercise 5.1

Q1: Write the first five terms of the AP:

(a) 3,6,9,12,…3, 6, 9, 12, \dots3,6,9,12,…

Solution: This is an AP where the first term a=3a = 3a=3 and the common difference d=6−3=3d = 6 – 3 = 3d=6−3=3. The terms are:a1=3,a2=3+3=6,a3=6+3=9,a4=9+3=12,a5=12+3=15a_1 = 3, \quad a_2 = 3 + 3 = 6, \quad a_3 = 6 + 3 = 9, \quad a_4 = 9 + 3 = 12, \quad a_5 = 12 + 3 = 15a1​=3,a2​=3+3=6,a3​=6+3=9,a4​=9+3=12,a5​=12+3=15

So, the first five terms are:
Answer: 3,6,9,12,153, 6, 9, 12, 153,6,9,12,15.

(b) 17,14,11,8,…17, 14, 11, 8, \dots17,14,11,8,…

Solution: Here, the first term a=17a = 17a=17 and the common difference d=14−17=−3d = 14 – 17 = -3d=14−17=−3. The terms are:a1=17,a2=17−3=14,a3=14−3=11,a4=11−3=8,a5=8−3=5a_1 = 17, \quad a_2 = 17 – 3 = 14, \quad a_3 = 14 – 3 = 11, \quad a_4 = 11 – 3 = 8, \quad a_5 = 8 – 3 = 5a1​=17,a2​=17−3=14,a3​=14−3=11,a4​=11−3=8,a5​=8−3=5

So, the first five terms are:
Answer: 17,14,11,8,517, 14, 11, 8, 517,14,11,8,5.

Q2: Find the nth term of the AP:
3,6,9,12,…3, 6, 9, 12, \dots3,6,9,12,…

Solution: The nth term of an AP is given by the formula:an=a+(n−1)×da_n = a + (n-1) \times dan​=a+(n−1)×d

Here, a=3a = 3a=3 and d=3d = 3d=3. So, the nth term is:an=3+(n−1)×3=3+3n−3=3na_n = 3 + (n-1) \times 3 = 3 + 3n – 3 = 3nan​=3+(n−1)×3=3+3n−3=3n

Answer: an=3na_n = 3nan​=3n.

Q3: Find the 10th term of the AP:
3,6,9,12,…3, 6, 9, 12, \dots3,6,9,12,…

Solution: We already know the formula for the nth term of the AP:an=3na_n = 3nan​=3n

For the 10th term:a10=3×10=30a_{10} = 3 \times 10 = 30a10​=3×10=30

Answer: The 10th term is 30.

Q4: Find the 20th term of the AP:
17,14,11,8,…17, 14, 11, 8, \dots17,14,11,8,…

Solution: Here, a=17a = 17a=17 and d=−3d = -3d=−3. Using the nth term formula:an=17+(n−1)×(−3)a_n = 17 + (n – 1) \times (-3)an​=17+(n−1)×(−3)

For the 20th term:a20=17+(20−1)×(−3)=17+19×(−3)=17−57=−40a_{20} = 17 + (20 – 1) \times (-3) = 17 + 19 \times (-3) = 17 – 57 = -40a20​=17+(20−1)×(−3)=17+19×(−3)=17−57=−40

Answer: The 20th term is −40-40−40.

Exercise 5.2

Q1: Find the sum of the first n terms of the AP:
2,6,10,14,…2, 6, 10, 14, \dots2,6,10,14,…

Solution: The sum of the first n terms of an AP is given by the formula:Sn=n2×(2a+(n−1)×d)S_n = \frac{n}{2} \times (2a + (n-1) \times d)Sn​=2n​×(2a+(n−1)×d)

Here, a=2a = 2a=2, d=4d = 4d=4. So, the sum of the first n terms is:Sn=n2×(2×2+(n−1)×4)=n2×(4+4n−4)=n2×4n=2n2S_n = \frac{n}{2} \times (2 \times 2 + (n-1) \times 4) = \frac{n}{2} \times (4 + 4n – 4) = \frac{n}{2} \times 4n = 2n^2Sn​=2n​×(2×2+(n−1)×4)=2n​×(4+4n−4)=2n​×4n=2n2

Answer: The sum of the first n terms is Sn=2n2S_n = 2n^2Sn​=2n2.

Q2: Find the sum of the first 15 terms of the AP:
3,7,11,15,…3, 7, 11, 15, \dots3,7,11,15,…

Solution: Here, a=3a = 3a=3, d=4d = 4d=4, and we need to find the sum of the first 15 terms, S15S_{15}S15​.

Using the formula:Sn=n2×(2a+(n−1)×d)S_n = \frac{n}{2} \times (2a + (n-1) \times d)Sn​=2n​×(2a+(n−1)×d)

Substitute the values:S15=152×(2×3+(15−1)×4)=152×(6+56)=152×62=15×31=465S_{15} = \frac{15}{2} \times (2 \times 3 + (15-1) \times 4) = \frac{15}{2} \times (6 + 56) = \frac{15}{2} \times 62 = 15 \times 31 = 465S15​=215​×(2×3+(15−1)×4)=215​×(6+56)=215​×62=15×31=465

Answer: The sum of the first 15 terms is 465.

Q3: Find the sum of the first 20 terms of the AP:
10,8,6,4,…10, 8, 6, 4, \dots10,8,6,4,…

Solution: Here, a=10a = 10a=10, d=−2d = -2d=−2, and we need to find the sum of the first 20 terms, S20S_{20}S20​.

Using the formula:Sn=n2×(2a+(n−1)×d)S_n = \frac{n}{2} \times (2a + (n-1) \times d)Sn​=2n​×(2a+(n−1)×d)

Substitute the values:S20=202×(2×10+(20−1)×(−2))=10×(20+19×(−2))=10×(20−38)=10×(−18)=−180S_{20} = \frac{20}{2} \times (2 \times 10 + (20-1) \times (-2)) = 10 \times (20 + 19 \times (-2)) = 10 \times (20 – 38) = 10 \times (-18) = -180S20​=220​×(2×10+(20−1)×(−2))=10×(20+19×(−2))=10×(20−38)=10×(−18)=−180

Answer: The sum of the first 20 terms is −180-180−180.

Exercise 5.3

Q1: The 10th term of an AP is 37, and the 15th term is 57. Find the first term and the common difference.

Solution: We are given:a10=37,a15=57a_{10} = 37, \quad a_{15} = 57a10​=37,a15​=57

The nth term formula is:an=a+(n−1)×da_n = a + (n – 1) \times dan​=a+(n−1)×d

For the 10th term:a10=a+9d=37(1)a_{10} = a + 9d = 37 \quad \text{(1)}a10​=a+9d=37(1)

For the 15th term:a15=a+14d=57(2)a_{15} = a + 14d = 57 \quad \text{(2)}a15​=a+14d=57(2)

Now, subtract equation (1) from equation (2):(a+14d)−(a+9d)=57−37(a + 14d) – (a + 9d) = 57 – 37(a+14d)−(a+9d)=57−37 5d=20⇒d=45d = 20 \quad \Rightarrow \quad d = 45d=20⇒d=4

Now substitute d=4d = 4d=4 into equation (1):a+9×4=37⇒a+36=37⇒a=1a + 9 \times 4 = 37 \quad \Rightarrow \quad a + 36 = 37 \quad \Rightarrow \quad a = 1a+9×4=37⇒a+36=37⇒a=1

Answer: The first term is 1 and the common difference is 4.

Q2: Find the sum of the first 20 terms of the AP:
7,11,15,19,…7, 11, 15, 19, \dots7,11,15,19,…

Solution: Here, a=7a = 7a=7, d=4d = 4d=4, and n=20n = 20n=20. Using the formula for the sum of the first n terms:Sn=n2×(2a+(n−1)×d)S_n = \frac{n}{2} \times (2a + (n-1) \times d)Sn​=2n​×(2a+(n−1)×d)

Substitute the values:S20=202×(2×7+(20−1)×4)=10×(14+76)=10×90=900S_{20} = \frac{20}{2} \times (2 \times 7 + (20-1) \times 4) = 10 \times (14 + 76) = 10 \times 90 = 900S20​=220​×(2×7+(20−1)×4)=10×(14+76)=10×90=900

Answer: The sum of the first 20 terms is 900.

These solutions cover the exercises related to Arithmetic Progressions in Class 10.

Additional Questions on Arithmetic Progressions (AP)

  1. Find the 8th term of the AP:
    5,8,11,14,…5, 8, 11, 14, \dots5,8,11,14,…
  2. Find the sum of the first 12 terms of the AP:
    7,10,13,16,…7, 10, 13, 16, \dots7,10,13,16,…
  3. The 5th term of an AP is 18, and the 10th term is 33. Find the first term and the common difference.
  4. Find the 20th term of the AP:
    2,6,10,14,…2, 6, 10, 14, \dots2,6,10,14,…
  5. The sum of the first 15 terms of an AP is 750. If the common difference is 5, find the first term.
  6. Find the common difference of the AP where the 6th term is 25 and the 10th term is 45.
  7. Find the sum of the first 50 terms of the AP:
    3,7,11,15,…3, 7, 11, 15, \dots3,7,11,15,…
  8. The nth term of an AP is given by an=5n−4a_n = 5n – 4an​=5n−4. Find the 15th term.
  9. In an AP, the first term is 2 and the 6th term is 20. Find the common difference.
  10. Find the sum of the first 25 terms of the AP:
    12,17,22,27,…12, 17, 22, 27, \dots12,17,22,27,…
  11. Find the 30th term of the AP:
    −3,1,5,9,…-3, 1, 5, 9, \dots−3,1,5,9,…
  12. The sum of the first 10 terms of an AP is 120. If the first term is 8, find the common difference.
  13. Find the 50th term of the AP:
    10,18,26,34,…10, 18, 26, 34, \dots10,18,26,34,…
  14. In an AP, the sum of the first 20 terms is 200. If the common difference is 3, find the first term.
  15. Find the sum of the first 40 terms of the AP:
    4,9,14,19,…4, 9, 14, 19, \dots4,9,14,19,…
  16. If the 7th term of an AP is 42 and the 13th term is 72, find the first term and the common difference.
  17. The sum of the first 30 terms of an AP is 1200. Find the first term if the common difference is 4.
  18. Find the sum of the first 15 terms of the AP:
    −1,3,7,11,…-1, 3, 7, 11, \dots−1,3,7,11,…
  19. The 20th term of an AP is 96, and the first term is 8. Find the common difference.
  20. Find the first term and the common difference of the AP where the 5th term is 22 and the 15th term is 52.

Solutions for Additional Questions:

  1. Find the 8th term of the AP:
    5,8,11,14,…5, 8, 11, 14, \dots5,8,11,14,…
    Formula: an=a+(n−1)×da_n = a + (n-1) \times dan​=a+(n−1)×d
    a=5,d=3a = 5, d = 3a=5,d=3
    a8=5+(8−1)×3=5+21=26a_8 = 5 + (8-1) \times 3 = 5 + 21 = 26a8​=5+(8−1)×3=5+21=26
  2. Find the sum of the first 12 terms of the AP:
    7,10,13,16,…7, 10, 13, 16, \dots7,10,13,16,…
    Formula: Sn=n2×[2a+(n−1)×d]S_n = \frac{n}{2} \times [2a + (n-1) \times d]Sn​=2n​×[2a+(n−1)×d]
    a=7,d=3,n=12a = 7, d = 3, n = 12a=7,d=3,n=12
    S12=122×[2×7+(12−1)×3]=6×[14+33]=6×47=282S_{12} = \frac{12}{2} \times [2 \times 7 + (12-1) \times 3] = 6 \times [14 + 33] = 6 \times 47 = 282S12​=212​×[2×7+(12−1)×3]=6×[14+33]=6×47=282
  3. The 5th term of an AP is 18, and the 10th term is 33. Find the first term and the common difference.
    Use the nth term formula:
    a5=a+4d=18a_5 = a + 4d = 18a5​=a+4d=18
    a10=a+9d=33a_{10} = a + 9d = 33a10​=a+9d=33
    Solve the system of equations:
    a+4d=18a + 4d = 18a+4d=18
    a+9d=33a + 9d = 33a+9d=33
    Subtracting the two equations:
    5d=15⇒d=35d = 15 \quad \Rightarrow \quad d = 35d=15⇒d=3
    Substitute d=3d = 3d=3 in a+4d=18a + 4d = 18a+4d=18:
    a+12=18⇒a=6a + 12 = 18 \quad \Rightarrow \quad a = 6a+12=18⇒a=6
  4. Find the 20th term of the AP:
    2,6,10,14,…2, 6, 10, 14, \dots2,6,10,14,…
    Formula: an=a+(n−1)×da_n = a + (n-1) \times dan​=a+(n−1)×d
    a=2,d=4a = 2, d = 4a=2,d=4
    a20=2+(20−1)×4=2+76=78a_{20} = 2 + (20-1) \times 4 = 2 + 76 = 78a20​=2+(20−1)×4=2+76=78
  5. The sum of the first 15 terms of an AP is 750. If the common difference is 5, find the first term.
    Use the sum formula:
    Sn=n2×[2a+(n−1)×d]S_n = \frac{n}{2} \times [2a + (n-1) \times d]Sn​=2n​×[2a+(n−1)×d]
    S15=750,n=15,d=5S_{15} = 750, n = 15, d = 5S15​=750,n=15,d=5
    750=152×[2a+14×5]750 = \frac{15}{2} \times [2a + 14 \times 5]750=215​×[2a+14×5]
    750=152×(2a+70)750 = \frac{15}{2} \times (2a + 70)750=215​×(2a+70)
    750=15×(a+35)750 = 15 \times (a + 35)750=15×(a+35)
    50=a+35⇒a=1550 = a + 35 \quad \Rightarrow \quad a = 1550=a+35⇒a=15
  6. Find the common difference of the AP where the 6th term is 25 and the 10th term is 45.
    Use the nth term formula:
    a6=a+5d=25a_6 = a + 5d = 25a6​=a+5d=25
    a10=a+9d=45a_{10} = a + 9d = 45a10​=a+9d=45
    Subtract the two equations:
    4d=20⇒d=54d = 20 \quad \Rightarrow \quad d = 54d=20⇒d=5
  7. Find the sum of the first 50 terms of the AP:
    3,7,11,15,…3, 7, 11, 15, \dots3,7,11,15,…
    Formula: Sn=n2×[2a+(n−1)×d]S_n = \frac{n}{2} \times [2a + (n-1) \times d]Sn​=2n​×[2a+(n−1)×d]
    a=3,d=4,n=50a = 3, d = 4, n = 50a=3,d=4,n=50
    S50=502×[2×3+(50−1)×4]=25×[6+196]=25×202=5050S_{50} = \frac{50}{2} \times [2 \times 3 + (50-1) \times 4] = 25 \times [6 + 196] = 25 \times 202 = 5050S50​=250​×[2×3+(50−1)×4]=25×[6+196]=25×202=5050
  8. The nth term of an AP is given by an=5n−4a_n = 5n – 4an​=5n−4. Find the 15th term.
    Substitute n=15n = 15n=15 into the formula:
    a15=5×15−4=75−4=71a_{15} = 5 \times 15 – 4 = 75 – 4 = 71a15​=5×15−4=75−4=71
  9. In an AP, the first term is 2 and the 6th term is 20. Find the common difference.
    Use the nth term formula:
    a6=a+5d=20a_6 = a + 5d = 20a6​=a+5d=20
    2+5d=20⇒5d=18⇒d=185=3.62 + 5d = 20 \quad \Rightarrow \quad 5d = 18 \quad \Rightarrow \quad d = \frac{18}{5} = 3.62+5d=20⇒5d=18⇒d=518​=3.6
  1. Find the sum of the first 25 terms of the AP:
    12,17,22,27,…12, 17, 22, 27, \dots12,17,22,27,…

Solution:

We are given:

  • First term a=12a = 12a=12
  • Common difference d=5d = 5d=5
  • Number of terms n=25n = 25n=25

Use the sum formula:Sn=n2×[2a+(n−1)×d]S_n = \frac{n}{2} \times [2a + (n-1) \times d]Sn​=2n​×[2a+(n−1)×d]

Substitute the values:S25=252×[2×12+(25−1)×5]S_{25} = \frac{25}{2} \times [2 \times 12 + (25-1) \times 5]S25​=225​×[2×12+(25−1)×5]S25=252×[24+120]=252×144=25×72=1800S_{25} = \frac{25}{2} \times [24 + 120] = \frac{25}{2} \times 144 = 25 \times 72 = 1800S25​=225​×[24+120]=225​×144=25×72=1800

Answer: The sum of the first 25 terms is 1800.

  1. Find the 30th term of the AP:
    −3,1,5,9,…-3, 1, 5, 9, \dots−3,1,5,9,…

Solution:

We are given:

  • First term a=−3a = -3a=−3
  • Common difference d=4d = 4d=4

Use the nth term formula:an=a+(n−1)×da_n = a + (n-1) \times dan​=a+(n−1)×d

Substitute n=30n = 30n=30:a30=−3+(30−1)×4=−3+29×4=−3+116=113a_{30} = -3 + (30-1) \times 4 = -3 + 29 \times 4 = -3 + 116 = 113a30​=−3+(30−1)×4=−3+29×4=−3+116=113

Answer: The 30th term is 113.

  1. The sum of the first 10 terms of an A P is 120. If the first term is 8, find the common difference.

Solution:

We are given:

  • Sum of first 10 terms S10=120S_{10} = 120S10​=120
  • First term a=8a = 8a=8
  • Number of terms n=10n = 10n=10

Use the sum formula:Sn=n2×[2a+(n−1)×d]S_n = \frac{n}{2} \times [2a + (n-1) \times d]Sn​=2n​×[2a+(n−1)×d]

Substitute the given values:120=102×[2×8+(10−1)×d]120 = \frac{10}{2} \times [2 \times 8 + (10-1) \times d]120=210​×[2×8+(10−1)×d]120=5×[16+9d]120 = 5 \times [16 + 9d]120=5×[16+9d]120=80+45d120 = 80 + 45d120=80+45d40=45d⇒d=4045=8940 = 45d \quad \Rightarrow \quad d = \frac{40}{45} = \frac{8}{9}40=45d⇒d=4540​=98​

Answer: The common difference is 89\frac{8}{9}98​.

  1. Find the 50th term of the AP:
    10,18,26,34,…10, 18, 26, 34, \dots10,18,26,34,…

Solution:

We are given:

  • First term a=10a = 10a=10
  • Common difference d=8d = 8d=8

Use the nth term formula:an=a+(n−1)×da_n = a + (n-1) \times dan​=a+(n−1)×d

Substitute n=50n = 50n=50:a50=10+(50−1)×8=10+49×8=10+392=402a_{50} = 10 + (50-1) \times 8 = 10 + 49 \times 8 = 10 + 392 = 402a50​=10+(50−1)×8=10+49×8=10+392=402

Answer: The 50th term is 402.

  1. In an AP, the sum of the first 20 terms is 200. If the common difference is 3, find the first term.

Solution:

We are given:

  • Sum of first 20 terms S20=200S_{20} = 200S20​=200
  • Common difference d=3d = 3d=3
  • Number of terms n=20n = 20n=20

Use the sum formula:Sn=n2×[2a+(n−1)×d]S_n = \frac{n}{2} \times [2a + (n-1) \times d]Sn​=2n​×[2a+(n−1)×d]

Substitute the given values:200=202×[2a+(20−1)×3]200 = \frac{20}{2} \times [2a + (20-1) \times 3]200=220​×[2a+(20−1)×3]200=10×[2a+57]200 = 10 \times [2a + 57]200=10×[2a+57]200=20a+570200 = 20a + 570200=20a+57020a=200−570=−37020a = 200 – 570 = -37020a=200−570=−370a=−37020=−18.5a = \frac{-370}{20} = -18.5a=20−370​=−18.5

Answer: The first term is −18.5-18.5−18.5.

  1. Find the sum of the first 40 terms of the AP:
    4,9,14,19,…4, 9, 14, 19, \dots4,9,14,19,…

Solution:

We are given:

  • First term a=4a = 4a=4
  • Common difference d=5d = 5d=5
  • Number of terms n=40n = 40n=40

Use the sum formula:Sn=n2×[2a+(n−1)×d]S_n = \frac{n}{2} \times [2a + (n-1) \times d]Sn​=2n​×[2a+(n−1)×d]

Substitute the given values:S40=402×[2×4+(40−1)×5]S_{40} = \frac{40}{2} \times [2 \times 4 + (40-1) \times 5]S40​=240​×[2×4+(40−1)×5]S40=20×[8+195]=20×203=4060S_{40} = 20 \times [8 + 195] = 20 \times 203 = 4060S40​=20×[8+195]=20×203=4060

Answer: The sum of the first 40 terms is 4060.

  1. If the 7th term of an AP is 42 and the 13th term is 72, find the first term and the common difference.

Solution:

We are given:

  • a7=a+6d=42a_7 = a + 6d = 42a7​=a+6d=42
  • a13=a+12d=72a_{13} = a + 12d = 72a13​=a+12d=72

Solve the system of equations:

  1. a+6d=42a + 6d = 42a+6d=42
  2. a+12d=72a + 12d = 72a+12d=72

Subtract the first equation from the second:(a+12d)−(a+6d)=72−42(a + 12d) – (a + 6d) = 72 – 42(a+12d)−(a+6d)=72−426d=30⇒d=56d = 30 \quad \Rightarrow \quad d = 56d=30⇒d=5

Substitute d=5d = 5d=5 into the first equation:a+6×5=42⇒a+30=42⇒a=12a + 6 \times 5 = 42 \quad \Rightarrow \quad a + 30 = 42 \quad \Rightarrow \quad a = 12a+6×5=42⇒a+30=42⇒a=12

Answer: The first term is 12 and the common difference is 5.

  1. The sum of the first 30 terms of an AP is 1200. Find the first term if the common difference is 4.

Solution:

We are given:

  • Sum of first 30 terms S30=1200S_{30} = 1200S30​=1200
  • Common difference d=4d = 4d=4
  • Number of terms n=30n = 30n=30

Use the sum formula:Sn=n2×[2a+(n−1)×d]S_n = \frac{n}{2} \times [2a + (n-1) \times d]Sn​=2n​×[2a+(n−1)×d]

Substitute the given values:1200=302×[2a+(30−1)×4]1200 = \frac{30}{2} \times [2a + (30-1) \times 4]1200=230​×[2a+(30−1)×4]1200=15×[2a+116]1200 = 15 \times [2a + 116]1200=15×[2a+116]1200=30a+17401200 = 30a + 17401200=30a+174030a=1200−1740=−54030a = 1200 – 1740 = -54030a=1200−1740=−540a=−54030=−18a = \frac{-540}{30} = -18a=30−540​=−18

Answer: The first term is −18-18−18.

  1. Find the sum of the first 15 terms of the AP:
    −1,3,7,11,…-1, 3, 7, 11, \dots−1,3,7,11,…

Solution:

We are given:

  • First term a=−1a = -1a=−1
  • Common difference d=4d = 4d=4
  • Number of terms n=15n = 15n=15

Use the sum formula:Sn=n2×[2a+(n−1)×d]S_n = \frac{n}{2} \times [2a + (n-1) \times d]Sn​=2n​×[2a+(n−1)×d]

Substitute the given values:S15=152×[2×(−1)+(15−1)×4]S_{15} = \frac{15}{2} \times [2 \times (-1) + (15-1) \times 4]S15​=215​×[2×(−1)+(15−1)×4]S15=152×[−2+56]=152×54=15×27=405S_{15} = \frac{15}{2} \times [-2 + 56] = \frac{15}{2} \times 54 = 15 \times 27 = 405S15​=215​×[−2+56]=215​×54=15×27=405

Answer: The sum of the first 15 terms is 405.

  1. The 20th term of an AP is 96, and the first term is 8. Find the common difference.

Solution:

We are given:

  • a20=96a_{20} = 96a20​=96
  • a=8a = 8a=8
  • n=20n = 20n=20

Use the nth term formula:an=a+(n−1)×da_n = a + (n-1) \times dan​=a+(n−1)×d

Substitute the values:96=8+(20−1)×d96 = 8 + (20-1) \times d96=8+(20−1)×d96=8+19d96 = 8 + 19d96=8+19d88=19d⇒d=8819=4.6388 = 19d \quad \Rightarrow \quad d = \frac{88}{19} = 4.6388=19d⇒d=1988​=4.63

Answer: The common difference is approximately 4.63.

  1. Find the first term and the common difference of the AP where the 5th term is 22 and the 15th term is 52.

Solution:

We are given:

  • a5=a+4d=22a_5 = a + 4d = 22a5​=a+4d=22
  • a15=a+14d=52a_{15} = a + 14d = 52a15​=a+14d=52

Solve the system of equations:

  1. a+4d=22a + 4d = 22a+4d=22
  2. a+14d=52a + 14d = 52a+14d=52

Subtract the first equation from the second:(a+14d)−(a+4d)=52−22(a + 14d) – (a + 4d) = 52 – 22(a+14d)−(a+4d)=52−2210d=30⇒d=310d = 30 \quad \Rightarrow \quad d = 310d=30⇒d=3

Substitute d=3d = 3d=3 into the first equation:a+4×3=22⇒a+12=22⇒a=10a + 4 \times 3 = 22 \quad \Rightarrow \quad a + 12 = 22 \quad \Rightarrow \quad a = 10a+4×3=22⇒a+12=22⇒a=10

Answer: The first term is 10 and the common difference is 3.

These are the solutions for all the additional questions. Let me know if you need more clarification or if you’d like additional questions!

Also Read: Class 9 Chapter 8 Maths “Statistics” Worksheets

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