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NCERT Exemplar · Class 10 Mathematics Arithmetic Progressions

71 questions · 71 still being checked

EXERCISE 5.3 11–20 (part 5 of 8)

  1. Exercise 11

    Determine k\displaystyle k so that k2+4k+8,2k2+3k+6,3k2+4k+4\displaystyle k^2+4 k+8,2 k^2+3 k+6,3 k^2+4 k+4 are three consecutive terms of an AP.

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    NCERT’s answer
    \(\displaystyle k=0\)
    For an AP, twice the middle term equals the sum of the outer two. \[2(2k^2+3k+6) = (k^2+4k+8)+(3k^2+4k+4) \] \[4k^2+6k+12 = 4k^2+8k+12 \] \[6k = 8k \implies k = 0 \]Answer: \(\displaystyle k=0\).

  2. Exercise 12

    Split 207\displaystyle 207 into three parts such that these are in AP and the product of the two smaller parts is 4623.

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    $\displaystyle 67$, $\displaystyle 69$, $\displaystyle 71$
    Let the parts be \(\displaystyle a-d,\ a,\ a+d\). \[(a-d)+a+(a+d) = 207 \implies 3a = 207 \implies a = 69 \] \[(a-d)\,a = 4623 \implies 69(69-d) = 4623 \] \[69-d = 67 \implies d = 2 \]Answer: \(\displaystyle 67,\ 69,\ 71\).

  3. Exercise 13

    The angles of a triangle are in AP. The greatest angle is twice the least. Find all the angles of the triangle.

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    $\displaystyle 40$°, $\displaystyle 60$°, $\displaystyle 80$°
    Let the angles be \(\displaystyle a-d,\ a,\ a+d\). \[(a-d)+a+(a+d) = 180 \implies 3a = 180 \implies a = 60 \] \[a+d = 2(a-d) \implies 3d = a = 60 \implies d = 20 \]Answer: \(\displaystyle 40^\circ,\ 60^\circ,\ 80^\circ\).

  4. Exercise 14

    If the n\displaystyle nth terms of the two APs: 9,7,5,…\displaystyle 9,7,5, \ldots and 24,21,18,…\displaystyle 24,21,18, \ldots are the same, find the value of n\displaystyle n. Also find that term.

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    \(\displaystyle 16^{\text {th }}\) term; -$\displaystyle 21$
    AP\(\displaystyle _1\): \(\displaystyle a=9,\ d=-2\); AP\(\displaystyle _2\): \(\displaystyle a=24,\ d=-3\). \[a_n^{(1)} = 9-2(n-1) = 11-2n \] \[a_n^{(2)} = 24-3(n-1) = 27-3n \] \[11-2n = 27-3n \implies n = 16 \] \[a_{16} = 11-2(16) = -21 \]Answer: \(\displaystyle n=16\); the term is \(\displaystyle -21\).

  5. Exercise 15

    If sum of the 3rd \displaystyle 3^{\text {rd }} and the 8th \displaystyle 8^{\text {th }} terms of an AP is 7\displaystyle 7 and the sum of the 7th \displaystyle 7^{\text {th }} and the 14th \displaystyle 14^{\text {th }} terms is -3\displaystyle 3, find the 10th \displaystyle 10^{\text {th }} term.

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    NCERT’s answer
    -$\displaystyle 1$
    \[a_3+a_8=(a+2d)+(a+7d)=2a+9d=7 \] \[a_7+a_{14}=(a+6d)+(a+13d)=2a+19d=-3 \] Subtracting, \[10d=-10 \quad\Rightarrow\quad d=-1 \] \[2a+9(-1)=7 \quad\Rightarrow\quad a=8 \] \[a_{10}=a+9d=8+9(-1)=-1 \] Answer: \(\displaystyle a_{10}=-1\)

  6. Exercise 16

    Find the 12th \displaystyle 12^{\text {th }} term from the end of the AP: −2,−4,−6,…,−100\displaystyle -2,-4,-6, \ldots,-100.

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    -$\displaystyle 78$
    Reversing the AP swaps first and last term and flips the sign of \(\displaystyle d\). \[l=-100,\quad d=-2 \] \[n^{\text{th}}\text{ term from the end}=l-(n-1)d \] \[12^{\text{th}}\text{ term from the end}=-100-11(-2)=-100+22=-78 \] Answer: \(\displaystyle -78\)

  7. Exercise 17

    Which term of the AP: 53\displaystyle 53, 48\displaystyle 48, 43\displaystyle 43,... is the first negative term?

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    12th term
    \[a=53,\quad d=48-53=-5 \] \[a_n=53-5(n-1)<0 \] \[5(n-1)>53 \quad\Rightarrow\quad n-1>10.6 \quad\Rightarrow\quad n\ge 12 \] \[a_{12}=53-5(11)=-2 \] Answer: the \(\displaystyle 12^{\text{th}}\) term \(\displaystyle (a_{12}=-2)\)

  8. Exercise 18

    How many numbers lie between 10\displaystyle 10 and 300\displaystyle 300, which when divided by 4\displaystyle 4 leave a remainder 3\displaystyle 3?

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    $\displaystyle 73$
    Numbers leaving remainder \(\displaystyle 3\) on division by \(\displaystyle 4\) form an AP. \[a=11,\quad d=4,\quad l=299 \] \[l=a+(n-1)d \] \[299=11+4(n-1) \quad\Rightarrow\quad n-1=72 \quad\Rightarrow\quad n=73 \] Answer: \(\displaystyle 73\)

  9. Exercise 19

    Find the sum of the two middle most terms of the AP: −43,−1,−23,…,413\displaystyle -\frac{4}{3},-1,-\frac{2}{3}, \ldots, 4 \frac{1}{3}.

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    $\displaystyle 3$
    \[a=-\frac{4}{3},\quad d=-1-\left(-\frac{4}{3}\right)=\frac{1}{3},\quad l=4\frac{1}{3}=\frac{13}{3} \] \[a+(n-1)d=l \quad\Rightarrow\quad -\frac{4}{3}+\frac{n-1}{3}=\frac{13}{3} \quad\Rightarrow\quad n=18 \] \(\displaystyle n\) is even, so the two middle terms are the \(\displaystyle 9^{\text{th}}\) and \(\displaystyle 10^{\text{th}}\): \[a_9=a+8d=-\frac{4}{3}+\frac{8}{3}=\frac{4}{3},\qquad a_{10}=a+9d=-\frac{4}{3}+3=\frac{5}{3} \] \[a_9+a_{10}=\frac{4}{3}+\frac{5}{3}=3 \] Answer: \(\displaystyle 3\)

  10. Exercise 20

    The first term of an AP is -5\displaystyle 5 and the last term is 45. If the sum of the terms of the AP is 120\displaystyle 120, then find the number of terms and the common difference.

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    \(\displaystyle n=6, d=10\)
    \[a=-5,\quad l=45,\quad S_n=120 \] \[S_n=\frac{n}{2}(a+l) \quad\Rightarrow\quad 120=\frac{n}{2}(40) \quad\Rightarrow\quad n=6 \] \[l=a+(n-1)d \quad\Rightarrow\quad 45=-5+5d \quad\Rightarrow\quad d=10 \] Answer: \(\displaystyle n=6,\ d=10\)