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NCERT Exemplar · Class 10 Mathematics Arithmetic Progressions

71 questions · 71 still being checked

EXERCISE 5.3 21–30 (part 6 of 8)

  1. Exercise 21

    Find the sum:
    (i)
    1+(−2)+(−5)+(−8)+…+(−236)\displaystyle 1+(-2)+(-5)+(-8)+\ldots+(-236)
    (ii)
    (4−1n)+(4−2n)+(4−3n)+…\displaystyle \left(4-\frac{1}{n}\right)+\left(4-\frac{2}{n}\right)+\left(4-\frac{3}{n}\right)+\ldots upto n\displaystyle n terms
    (iii)
    a−ba+b+3a−2ba+b+5a−3ba+b+⋯\displaystyle \frac{a-b}{a+b}+\frac{3 a-2 b}{a+b}+\frac{5 a-3 b}{a+b}+\cdots to 11\displaystyle 11 terms.

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    NCERT’s answer
    (i)
    -$\displaystyle 9400$
    (ii)
    \(\displaystyle \frac{7 n-1}{2}\)
    (iii)
    \(\displaystyle \frac{11(11 a-6 b)}{a+b}\)
    (i)
    \[a=1,\quad d=-2-1=-3,\quad l=-236 \]
    \[l=a+(n-1)d \quad\Rightarrow\quad -236=1-3(n-1) \quad\Rightarrow\quad n=80 \]
    \[S_{80}=\frac{80}{2}(1-236)=40(-235)=-9400 \]
    (ii)
    \[a_1=4-\frac{1}{n},\quad d=-\frac{1}{n},\quad a_n=4-\frac{n}{n}=3 \]
    \[S_n=\frac{n}{2}(a_1+a_n)=\frac{n}{2}\left(4-\frac{1}{n}+3\right)=\frac{7n-1}{2} \]
    (iii)
    The \(\displaystyle k^{\text{th}}\) term is \(\displaystyle \dfrac{(2k-1)a-kb}{a+b}\).
    \[\sum_{k=1}^{11}\big[(2k-1)a-kb\big]=a\sum_{k=1}^{11}(2k-1)-b\sum_{k=1}^{11}k=121a-66b \]
    \[S_{11}=\frac{121a-66b}{a+b}=\frac{11(11a-6b)}{a+b} \]
    Answer: (i) \(\displaystyle -9400\) (ii) \(\displaystyle \dfrac{7n-1}{2}\) (iii) \(\displaystyle \dfrac{11(11a-6b)}{a+b}\)

  2. Exercise 22

    Which term of the AP: −2,−7,−12,…\displaystyle -2,-7,-12, \ldots will be −77\displaystyle -77? Find the sum of this AP upto the term −77\displaystyle -77.

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    NCERT’s answer
    \(\displaystyle 16^{\text {th }}\) term; -$\displaystyle 632$
    \[a = -2, \quad d = -7-(-2) = -5 \] \[a_n = a+(n-1)d \] \[-77 = -2+(n-1)(-5) \] \[n-1 = 15 \quad \Rightarrow \quad n = 16 \] \[S_{16} = \frac{16}{2}\big[2(-2)+15(-5)\big] = 8(-79) = -632 \] Answer: the \(\displaystyle 16^{\text{th}}\) term is \(\displaystyle -77\); \(\displaystyle S_{16} = -632\).

  3. Exercise 23

    If an=3−4n\displaystyle a_n=3-4 n, show that a1,a2,a3,…\displaystyle a_1, a_2, a_3, \ldots form an AP. Also find S20\displaystyle \mathrm{S}_{20}.

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    NCERT’s answer
    -$\displaystyle 780$
    \[a_n = 3-4n \] \[a_n - a_{n-1} = (3-4n)-\big(3-4(n-1)\big) = -4 \] Constant, independent of \(\displaystyle n\), so the terms form an AP with \(\displaystyle d=-4\). \[a_1 = 3-4(1) = -1 \] \[S_{20} = \frac{20}{2}\big[2(-1)+19(-4)\big] = 10(-78) = -780 \] Answer: AP with \(\displaystyle d=-4\); \(\displaystyle S_{20} = -780\).

  4. Exercise 24

    In an AP, if Sn=n(4n+1)\displaystyle \mathrm{S}_n=n(4 n+1), find the AP.

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    NCERT’s answer
    $\displaystyle 5$, $\displaystyle 13$, $\displaystyle 21$, ---
    \[a_1 = S_1 = 1\big(4(1)+1\big) = 5 \] \[S_2 = 2\big(4(2)+1\big) = 18, \quad a_2 = S_2-S_1 = 13 \] \[d = a_2-a_1 = 8 \] \[a_n = S_n-S_{n-1} = (4n^2+n)-\big(4(n-1)^2+(n-1)\big) = 8n-3 \] Answer: \(\displaystyle 5, 13, 21, 29, \ldots\) (\(\displaystyle a=5,\ d=8\)).

  5. Exercise 25

    In an AP, if Sn=3n2+5n\displaystyle \mathrm{S}_n=3 n^2+5 n and ak=164\displaystyle a_k=164, find the value of k\displaystyle k.

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    NCERT’s answer
    \(\displaystyle k=27\)
    \[S_n = 3n^2+5n \] \[a_n = S_n-S_{n-1} = (3n^2+5n)-\big(3(n-1)^2+5(n-1)\big) = 6n+2 \] \[a_k = 6k+2 = 164 \] \[6k = 162 \quad \Rightarrow \quad k = 27 \] Answer: \(\displaystyle k = 27\).

  6. Exercise 26

    If Sn\displaystyle \mathrm{S}_n denotes the sum of first n\displaystyle n terms of an AP, prove that S12=3(S8−S4)\mathrm{S}_{12}=3\left(\mathrm{S}_8-\mathrm{S}_4\right)

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    \[S_n = \frac{n}{2}\big[2a+(n-1)d\big] \] \[S_{12} = \frac{12}{2}\big[2a+11d\big] = 12a+66d \] \[S_{8} = \frac{8}{2}\big[2a+7d\big] = 8a+28d \] \[S_{4} = \frac{4}{2}\big[2a+3d\big] = 4a+6d \] \[S_8-S_4 = 4a+22d \] \[3(S_8-S_4) = 12a+66d = S_{12} \] Answer: \(\displaystyle S_{12} = 3(S_8-S_4)\).

  7. Exercise 27

    Find the sum of first 17\displaystyle 17 terms of an AP whose 4th \displaystyle 4^{\text {th }} and 9th \displaystyle 9^{\text {th }} terms are −15\displaystyle -15 and −30\displaystyle -30 respectively.

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    NCERT’s answer
    -$\displaystyle 510$
    \[a+3d = -15 \] \[a+8d = -30 \] \[5d = -15 \quad \Rightarrow \quad d = -3 \] \[a = -15-3(-3) = -6 \] \[S_{17} = \frac{17}{2}\big[2(-6)+16(-3)\big] = 17(-30) = -510 \] Answer: \(\displaystyle S_{17} = -510\).

  8. Exercise 28

    If sum of first 6\displaystyle 6 terms of an AP is 36\displaystyle 36 and that of the first 16\displaystyle 16 terms is 256\displaystyle 256, find the sum of first 10\displaystyle 10 terms.

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    NCERT’s answer
    $\displaystyle 100$
    \[S_6 = 3\big[2a+5d\big] = 36 \quad \Rightarrow \quad 2a+5d=12 \] \[S_{16} = 8\big[2a+15d\big] = 256 \quad \Rightarrow \quad 2a+15d=32 \] \[10d = 20 \quad \Rightarrow \quad d=2, \quad a=1 \] \[S_{10} = 5\big[2(1)+9(2)\big] = 5(20) = 100 \] Answer: \(\displaystyle S_{10} = 100\).

  9. Exercise 29

    Find the sum of all the 11\displaystyle 11 terms of an AP whose middle most term is 30.

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    NCERT’s answer
    $\displaystyle 330$
    For $\displaystyle 11$ terms (odd count), the middle term is the 6th, and the sum equals \(\displaystyle n\) times the middle term. \[S_{11} = \frac{11}{2}(a_1+a_{11}) = 11 \times a_6 \] \[a_6 = 30 \] \[S_{11} = 11 \times 30 = 330 \] Answer: \(\displaystyle S_{11} = 330\)

  10. Exercise 30

    Find the sum of last ten terms of the AP: 8\displaystyle 8, 10\displaystyle 10, 12\displaystyle 12,---, 126.

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    NCERT’s answer
    $\displaystyle 1170$
    The last ten terms form an AP starting at the last term $\displaystyle 126$ with common difference \(\displaystyle -2\). \[a = 126,\ d = -2,\ n = 10 \] \[S_{10} = \frac{10}{2}\big[2(126) + 9(-2)\big] = 5(252-18) \] \[S_{10} = 1170 \] Answer: \(\displaystyle 1170\)