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NCERT Exemplar · Class 10 Mathematics Arithmetic Progressions

71 questions · 71 still being checked

EXERCISE 5.1 11–18 (part 2 of 8)

  1. Choose the correct answer from the given four options:

    Exercise 11

    Two APs have the same common difference. The first term of one of these is -1\displaystyle 1 and that of the other is -8. Then the difference between their 4th \displaystyle 4^{\text {th }} terms is
    (A)
    -1\displaystyle 1 (B) -8\displaystyle 8 (C) 7\displaystyle 7 (D) -9\displaystyle 9

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    NCERT’s answer
    (C)
    (C) \(\displaystyle 7\)\[a_4 = a_1 + 3d,\quad a_4' = a_1' + 3d \] \[a_4 - a_4' = a_1 - a_1' \] \[= -1 - (-8) = 7 \]

  2. Exercise 12

    If 7\displaystyle 7 times the 7th \displaystyle 7^{\text {th }} term of an AP is equal to 11\displaystyle 11 times its 11th \displaystyle 11^{\text {th }} term, then its 18th term will be
    (A)
    7\displaystyle 7 (B) 11\displaystyle 11
    (C)
    18\displaystyle 18

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    NCERT’s answer
    (D)
    (D) \(\displaystyle 0\)\[7a_7 = 11a_{11} \] \[7(a+6d) = 11(a+10d) \] \[4a + 68d = 0 \implies a = -17d \] \[a_{18} = a + 17d = -17d + 17d = 0 \]

  3. Exercise 13

    The 4th \displaystyle 4^{\text {th }} term from the end of the AP: -11\displaystyle 11, -8\displaystyle 8, -5\displaystyle 5, ..., 49\displaystyle 49 is
    (A)
    37\displaystyle 37
    (B)
    40\displaystyle 40
    (C)
    43\displaystyle 43
    (D)
    58\displaystyle 58

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    NCERT’s answer
    (B)
    (B) \(\displaystyle 40\) \[a=-11,\quad d=3,\quad l=49 \] \[n\text{th term from end}=l-(n-1)d \] \[4\text{th term from end}=49-(4-1)(3)=40 \]

  4. Exercise 14

    The famous mathematician associated with finding the sum of the first 100\displaystyle 100 natural numbers is
    (A)
    Pythagoras
    (B)
    Newton
    (C)
    Gauss
    (D)
    Euclid

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    NCERT’s answer
    (C)
    (C) Gauss \[1+2+\cdots+100=\frac{100(101)}{2}=5050 \] He paired terms symmetrically as a child.

  5. Exercise 15

    If the first term of an AP is -5\displaystyle 5 and the common difference is 2\displaystyle 2, then the sum of the first 6\displaystyle 6 terms is
    (A)
    0\displaystyle 0 (B) 5\displaystyle 5 (C) 6\displaystyle 6 (D) 15\displaystyle 15

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    NCERT’s answer
    (A)
    (A) \(\displaystyle 0\) \[a=-5,\quad d=2,\quad n=6 \] \[S_n=\frac{n}{2}\left[2a+(n-1)d\right] \] \[S_6=3\left[-10+10\right]=0 \]

  6. Exercise 16

    The sum of first 16\displaystyle 16 terms of the AP: 10\displaystyle 10, 6\displaystyle 6, 2\displaystyle 2,... is
    (A)
    -320\displaystyle 320
    (B)
    320\displaystyle 320
    (C)
    -352\displaystyle 352
    (D)
    -400\displaystyle 400

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    NCERT’s answer
    (A)
    (A) \(\displaystyle -320\) \[a=10,\quad d=-4,\quad n=16 \] \[S_{16}=\frac{16}{2}\left[2(10)+15(-4)\right] \] \[=8(20-60)=-320 \]

  7. Exercise 17

    In an AP if a=1,an=20\displaystyle a=1, a_n=20 and Sn=399\displaystyle \mathrm{S}_n=399, then n\displaystyle n is
    (A)
    19\displaystyle 19
    (B)
    21\displaystyle 21
    (C)
    38\displaystyle 38
    (D)
    42\displaystyle 42

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    NCERT’s answer
    (C)
    (C) \(\displaystyle 38\) \[S_n=\frac{n}{2}(a+a_n) \] \[399=\frac{n}{2}(1+20) \] \[n=\frac{798}{21}=38 \]

  8. Exercise 18

    The sum of first five multiples of 3\displaystyle 3 is
    (A)
    45\displaystyle 45
    (B)
    55\displaystyle 55
    (C)
    65\displaystyle 65
    (D)
    75\displaystyle 75

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    NCERT’s answer
    (A)
    (A) \(\displaystyle 45\) \[3,6,9,12,15 \] \[S_5=\frac{5}{2}\left[2(3)+4(3)\right]=45 \]