Home/ jaanvi roy/Best Answers
  1. Asked: In:

    The houses of a row are numbered consecutively from 1 to 49. Show that there is a value of x such that the sum of the numbers of the houses preceding the house numbered x is equal to the sum of the numbers of the houses following it. Find this value of x.

    jaanvi roy

    Here, a = 1 and d = 1. Sum to n terms of an AP is given by S_n = n/2[ 2a + (n -1)d] The sum of the numbers of the houses preceding the house numbered x is equal to the sum of the numbers of the houses following it. Therefore Sx-₁ = S₄₉ - Sx ⇒ (x -1/2)[2a + (x - 1 - 1)d] ⇒ 49/2[2a + (49 - 1)d] - x/2[Read more

    Here, a = 1 and d = 1.
    Sum to n terms of an AP is given by
    S_n = n/2[ 2a + (n -1)d]
    The sum of the numbers of the houses preceding the house numbered x is equal to the sum of the numbers of the houses following it. Therefore
    Sx-₁ = S₄₉ – Sx
    ⇒ (x -1/2)[2a + (x – 1 – 1)d]
    ⇒ 49/2[2a + (49 – 1)d] – x/2[2a + (x -1)d]
    ⇒ x -1/2[2(1) + (x -2)(1)]
    ⇒ 49/2[2(1) + 48(1)] – x/2[(1) + (x – 1)(1)]
    ⇒ (x – 1)[x] = 49[50] – x[x + 1]
    ⇒ x² – x = 2450 – x² – x
    ⇒ 2x² = 2450 x² = 1225 ⇒ x = 35
    Hence, the value of x is 35.

    See less
    • 1
  2. Asked: In:

    Find the sum of the following APs: –37, –33, –29, . . ., to 12 terms.

    jaanvi roy

    A.P. : 37, 33, 29, ... Here, a = -37 and d = 33 - (-37) = 4. The sum of n terms of an AP is Given by S_n = n/2[2a + (n -1)d] ⇒ S₁₂ = 12/2[2(-37) + (12 -1)(4)] ⇒ S₁₂ = 6[- 74 + 44] = - 180

    A.P. : 37, 33, 29, …
    Here, a = -37 and d = 33 – (-37) = 4.
    The sum of n terms of an AP is Given by
    S_n = n/2[2a + (n -1)d]
    ⇒ S₁₂ = 12/2[2(-37) + (12 -1)(4)]
    ⇒ S₁₂ = 6[- 74 + 44] = – 180

    See less
    • 2
  3. Asked: In:

    If the sum of the first n terms of an AP is 4n – n², what is the first term (that is S1)? What is the sum of first two terms? What is the second term? Similarly, find the 3rd, the 10th and the nth terms.

    jaanvi roy

    The sum of n terms of an AP is given by S_n = 4n - n² Putting n = 1, we get First term = a₁ = S₁ = 4(1) - (1) ² = 3 Putting n = 2, we get sum of two terms = a₁ + a₂ = S₂ = 4(2) -(2)² = 4 ⇒ a₁ + a₂ = 4 ⇒ 3 + a₂ = 4 [∵ the first term a₁ = 3 ] ⇒ a₂ = 1 Hence, the second term is 1. Common difference d =Read more

    The sum of n terms of an AP is given by
    S_n = 4n – n²
    Putting n = 1, we get
    First term = a₁ = S₁ = 4(1) – (1) ² = 3
    Putting n = 2, we get
    sum of two terms = a₁ + a₂ = S₂ = 4(2) -(2)² = 4
    ⇒ a₁ + a₂ = 4
    ⇒ 3 + a₂ = 4 [∵ the first term a₁ = 3 ]
    ⇒ a₂ = 1
    Hence, the second term is 1.
    Common difference d = a₂ – a₁ = 1 – 3 = – 2
    Therefore, the tenth term = a_10 = a + 9d = 3 + 9(-2) = – 16
    Similarly, the nth term = a_n = a + (n -1)d = 3 + (n -1)(-2) = 5 – 2n

    See less
    • 1
  4. Asked: In:

    Which term of the AP : 121, 117, 113, . . ., is its first negative term?

    jaanvi roy

    Here, a = 121 and d = 117 - 121 = - 4. Let, a_n be the first negative term of this AP. ⇒ a_n < 0 ⇒ a + (n - 1)d < 0 ⇒ 121 + (n -1)(-4) < 0 ⇒ 121 - 4n + 4 < 0 ⇒ 125 125/4 ⇒ n > 31.25 ⇒ n = 32 Hence, 32nd term of the AP: 121, 117, 113 ... is its first negative term.

    Here, a = 121 and d = 117 – 121 = – 4.
    Let, a_n be the first negative term of this AP.
    ⇒ a_n < 0
    ⇒ a + (n – 1)d < 0
    ⇒ 121 + (n -1)(-4) < 0
    ⇒ 121 – 4n + 4 < 0
    ⇒ 125 125/4
    ⇒ n > 31.25
    ⇒ n = 32
    Hence, 32nd term of the AP: 121, 117, 113 … is its first negative term.

    See less
    • 1
  6. Asked: In:

    The sum of the third and the seventh terms of an AP is 6 and their product is 8. Find the sum of first sixteen terms of the AP.

    jaanvi roy

    Let, the first term = a and common difference = d the sum of the third and the seventh terms of the AP is 6, therefore a₃ + a₇ = 6 ⇒ a + 2d + a + 6d = 6 ⇒ 2a + 8d = 6 ⇒ a + 4d = 3 ⇒ a = 3 - 4d ...(1) The product of the third and the seventh terms of the AP is 8, therefore (a₃)(a₇) = 8 ⇒ (a + 2d)(a +Read more

    Let, the first term = a and common difference = d
    the sum of the third and the seventh terms of the AP is 6, therefore
    a₃ + a₇ = 6
    ⇒ a + 2d + a + 6d = 6
    ⇒ 2a + 8d = 6
    ⇒ a + 4d = 3
    ⇒ a = 3 – 4d …(1)
    The product of the third and the seventh terms of the AP is 8, therefore
    (a₃)(a₇) = 8
    ⇒ (a + 2d)(a + 6d) = 8
    Putting the value of a from the equation(1), we get
    (3 -2d)(3 +2d) = 8
    ⇒ 3² – 4d² = 8
    ⇒ 4d² = 1
    ⇒ d = ± 1/2
    if, d = 1/2,
    Putting the value of d in the equation (1), we get
    a = 3 – 4(1/2) = 1
    The sum of the 16 terms of this AP is given by
    S₁₆ = 16/2[2a + 16 – 1)d] = 8[2(1) + 15 (1/2)] = 76
    If, d = – 1/2,
    Putting the value of d in the equation (1), we get
    a = 3 – 4(-1/2) = 5
    The sum of the 16 terms of this AP is given by
    S₁₆ = 16/2[2a + 16 – 1)d] = 8 [2(5) + 15 (-1/2)] = 20
    Hence, the sum of the 16 terms of AP is 20 or 76.

    See less
    • 1
  7. Asked: In:

    Find the sum of the following APs: 1/15, 1/12, 1/10, …, to 11 terms.

    jaanvi roy

    A.P. 1/15, 1/12, 1/10, ... Here, a = 1/15 and d = 1/12 - (1/15) = 1/60. The sum of n terms of an AP is Given by Sn = n/2[2a + (n -1)d] The sum of n terms of an AP is given by S_n = n/2[2a + (n - 1)d ⇒ S₁₁ = 11/2[2(1/15) + (11 - 1) (1/60)] ⇒ S₁₁ = 11/2[2/15+ 1/6] = 11/2[9/30] = 33/20.

    A.P. 1/15, 1/12, 1/10, …
    Here, a = 1/15 and d = 1/12 – (1/15) = 1/60.
    The sum of n terms of an AP is Given by
    Sn = n/2[2a + (n -1)d]
    The sum of n terms of an AP is given by
    S_n = n/2[2a + (n – 1)d
    ⇒ S₁₁ = 11/2[2(1/15) + (11 – 1) (1/60)]
    ⇒ S₁₁ = 11/2[2/15+ 1/6] = 11/2[9/30] = 33/20.

    See less
    • 2
  8. Asked: In:

    Find the sum of the following APs: 0.6, 1.7, 2.8, . . ., to 100 terms.

    jaanvi roy

    A.P. : 0.6, 1.7, 2.8, ... Here, a = 0.6, and d = 1.7 - 0.6 = 1.1. The sum of n terms of an AP is Given by Sn = n/2[2a + (n -1)d] ⇒ S_100 = 100/2[2(0.6) + (100 -1)(1.1) ⇒ S_100 = 50[1.2 + 99 × 1.1] = 50[110.1] = 5505

    A.P. : 0.6, 1.7, 2.8, …
    Here, a = 0.6, and d = 1.7 – 0.6 = 1.1.
    The sum of n terms of an AP is Given by
    Sn = n/2[2a + (n -1)d]
    ⇒ S_100 = 100/2[2(0.6) + (100 -1)(1.1)
    ⇒ S_100 = 50[1.2 + 99 × 1.1] = 50[110.1] = 5505

    See less
    • 2
  9. Asked: In:

    In a school, students thought of planting trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant, will be the same as the class, in which they are studying, e.g., a section of Class I will plant 1 tree, a section of Class II will plant 2 trees and so on till Class XII. There are three sections of each class. How many trees will be planted by the students?

    jaanvi roy

    Each section of each class will plant tree = 3 × Class, therefore Total number of tree planted by class i = 3 × 1 = 3 Total number of tree planted by class ii = 3 × 2 = 6 Total number of tree planted by class iii = 3 × 3 = 9 Similary, the series of trees planted by classes are as follows: 3, 6, 9, .Read more

    Each section of each class will plant tree = 3 × Class, therefore
    Total number of tree planted by class i = 3 × 1 = 3
    Total number of tree planted by class ii = 3 × 2 = 6
    Total number of tree planted by class iii = 3 × 3 = 9
    Similary, the series of trees planted by classes are as follows: 3, 6, 9, …., 36
    Here, a = 3, d = 6 – 3 = 3 and n = 12.
    The sum of n terms of an AP is given by S_n = n/2[2a + (n -1)d]
    ⇒ S₁₂ = 12 /2[2(3) + (12 – 1)(3)]
    = 6[6 + 33] = 6(39) = 234
    Hence, the total number of tree planted by the students is 234.

    See less
    • 1
  11. Asked: In:

    Find the sums given below: –5 + (–8) + (–11) + . . . + (–230)

    jaanvi roy

    Here, a = - 5 and d = - 8 - (-5) = -3. Let, the nth term of the A.P. is - 230. Therefore, a_n = -230 ⇒ a + (n -1)d = - 230 ⇒ - 5 + (n - 1)(-3) = - 230 ⇒ (n - 1)(-3) = -225 ⇒ n - 1 = 75 ⇒ n = 76 The sum of n terms of an AP is given by S_n = n/2[a +l] ⇒ S₇₆ = 76/2[- 5 - 230] ⇒ S₇₆ = 76/2[- 235] = - 38Read more

    Here, a = – 5 and d = – 8 – (-5) = -3.
    Let, the nth term of the A.P. is – 230.
    Therefore, a_n = -230
    ⇒ a + (n -1)d = – 230
    ⇒ – 5 + (n – 1)(-3) = – 230
    ⇒ (n – 1)(-3) = -225
    ⇒ n – 1 = 75 ⇒ n = 76
    The sum of n terms of an AP is given by
    S_n = n/2[a +l]
    ⇒ S₇₆ = 76/2[- 5 – 230]
    ⇒ S₇₆ = 76/2[- 235]
    = – 38 × 235 = – 8930

    See less
    • 2
  12. Asked: In:

    Find the sums given below : 7 + 10(1/2) + 14 + … + 84

    jaanvi roy

    Here, a = 7 and d = 10(1/2) - 7 = 21/2 - 7 = 7/2. Let, the nth term of the A.P. is 84. Therefore, a_n = 84 ⇒ a + (n -1)d = 84 ⇒ 7 + (n - 1)(7/2) = 84 ⇒ (n - 1)(7/2) = 77 ⇒ n - 1 = 22 ⇒ n = 23 The sum of n terms of an AP is given by S_n = n/2[a +l] ⇒ S₂₃ = 23/2[7 + 84] ⇒ S₂₃ = 23/2[91] = 2093/2 = 104Read more

    Here, a = 7 and d = 10(1/2) – 7 = 21/2 – 7 = 7/2.
    Let, the nth term of the A.P. is 84.
    Therefore, a_n = 84
    ⇒ a + (n -1)d = 84
    ⇒ 7 + (n – 1)(7/2) = 84
    ⇒ (n – 1)(7/2) = 77
    ⇒ n – 1 = 22
    ⇒ n = 23
    The sum of n terms of an AP is given by
    S_n = n/2[a +l]
    ⇒ S₂₃ = 23/2[7 + 84]
    ⇒ S₂₃ = 23/2[91] = 2093/2 = 1046 (1/2)

    See less
    • 1