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NCERT Exemplar · Class 10 Mathematics Arithmetic Progressions

71 questions · 71 still being checked

EXERCISE 5.4 1–10 (part 8 of 8)

  1. Exercise 1

    The sum of the first five terms of an AP and the sum of the first seven terms of the same AP is 167. If the sum of the first ten terms of this AP is 235\displaystyle 235, find the sum of its first twenty terms.

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    NCERT’s answer
    $\displaystyle 970$
    Let \(\displaystyle a\), \(\displaystyle d\) be the first term and common difference. \[S_5+S_7=(5a+10d)+(7a+21d)=12a+31d=167 \] \[S_{10}=10a+45d=235 \implies 2a+9d=47 \] \[12a+31d=167,\quad 6(2a+9d)=12a+54d=282 \] \[23d=282-167=115 \implies d=5 \] \[2a+9(5)=47 \implies a=1 \] \[S_{20}=\frac{20}{2}\left[2(1)+19(5)\right]=10(97)=970 \]

  2. Exercise 2

    Find the
    (i)
    sum of those integers between 1\displaystyle 1 and 500\displaystyle 500 which are multiples of 2\displaystyle 2 as well as of 5.
    (ii)
    sum of those integers from 1\displaystyle 1 to 500\displaystyle 500 which are multiples of 2\displaystyle 2 as well as of 5\displaystyle 5 .
    (iii)
    sum of those integers from 1\displaystyle 1 to 500\displaystyle 500 which are multiples of 2\displaystyle 2 or 5.
    [Hint (iii) : These numbers will be : multiples of 2+\displaystyle 2+ multiples of 5−\displaystyle 5- multiples of 2\displaystyle 2 as well as of 5\displaystyle 5 ]

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    NCERT’s answer
    (i)
    $\displaystyle 12250$
    (ii)
    $\displaystyle 12750$
    (iii)
    $\displaystyle 75250$
    Multiples of $\displaystyle 2$ as well as of $\displaystyle 5$ are multiples of 10.
    (i)
    Between $\displaystyle 1$ and $\displaystyle 500$: \(\displaystyle 10,20,\dots,490\).
    \[n=49,\quad S=\frac{49}{2}(10+490)=12250 \]
    (ii)
    From $\displaystyle 1$ to $\displaystyle 500$: \(\displaystyle 10,20,\dots,500\).
    \[n=50,\quad S=\frac{50}{2}(10+500)=12750 \]
    (iii)
    From $\displaystyle 1$ to $\displaystyle 500$, multiples of $\displaystyle 2$ or $\displaystyle 5$:
    \[S_2=\frac{250}{2}(2+500)=62750 \]
    \[S_5=\frac{100}{2}(5+500)=25250 \]
    \[S_2+S_5-S_{10}=62750+25250-12750=75250 \]

  3. Exercise 3

    The eighth term of an AP is half its second term and the eleventh term exceeds one third of its fourth term by 1. Find the 15th \displaystyle 15^{\text {th }} term.

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    NCERT’s answer
    $\displaystyle 3$
    Let \(\displaystyle a\), \(\displaystyle d\) be the first term and common difference. \[a_8=\tfrac{1}{2}a_2 \implies a+7d=\tfrac{1}{2}(a+d) \implies a+13d=0 \] \[a_{11}=\tfrac{1}{3}a_4+1 \implies a+10d=\tfrac{1}{3}(a+3d)+1 \implies 2a+27d=3 \] \[a=-13d \implies -26d+27d=3 \implies d=3,\ a=-39 \] \[a_{15}=a+14d=-39+42=3 \]

  4. Exercise 4

    An AP consists of 37\displaystyle 37 terms. The sum of the three middle most terms is 225\displaystyle 225 and the sum of the last three is 429. Find the AP.

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    NCERT’s answer
    $\displaystyle 3$, $\displaystyle 7$, $\displaystyle 11$, $\displaystyle 15$, ---
    With $\displaystyle 37$ terms the three middle most are \(\displaystyle a_{18},a_{19},a_{20}\) and the last three are \(\displaystyle a_{35},a_{36},a_{37}\). \[a_{18}+a_{19}+a_{20}=3a_{19}=225 \implies a+18d=75 \] \[a_{35}+a_{36}+a_{37}=3a_{36}=429 \implies a+35d=143 \] \[17d=143-75=68 \implies d=4,\quad a=3 \] \[a_{37}=a+36d=3+144=147 \]

  5. Exercise 5

    Find the sum of the integers between 100\displaystyle 100 and 200\displaystyle 200 that are
    (i)
    divisible by 9\displaystyle 9
    (ii)
    not divisible by 9\displaystyle 9
    [Hint (ii) : These numbers will be : Total numbers - Total numbers divisible by 9\displaystyle 9]

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    NCERT’s answer
    (i)
    $\displaystyle 1683$
    (ii)
    $\displaystyle 13167$
    (i)
    Multiples of $\displaystyle 9$ between $\displaystyle 100$ and $\displaystyle 200$ run \(\displaystyle 108,117,\dots,198\).
    \[n=\frac{198-108}{9}+1=11,\quad S_9=\frac{11}{2}(108+198)=1683 \]
    (ii)
    Integers between $\displaystyle 100$ and $\displaystyle 200$ are \(\displaystyle 101,\dots,199\), a total of $\displaystyle 99$ numbers.
    \[S_{\text{total}}=\frac{99}{2}(101+199)=14850 \]
    \[S_{\text{total}}-S_9=14850-1683=13167 \]

  6. Exercise 6

    The ratio of the 11th \displaystyle 11^{\text {th }} term to the 18th \displaystyle 18^{\text {th }} term of an AP is 2:3\displaystyle 2: 3. Find the ratio of the 5th \displaystyle 5^{\text {th }} term to the 21st \displaystyle 21^{\text {st }} term, and also the ratio of the sum of the first five terms to the sum of the first 21\displaystyle 21 terms.

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    NCERT’s answer
    $\displaystyle 1$:$\displaystyle 3$; $\displaystyle 5$:$\displaystyle 49$
    Let first term \(\displaystyle a\) and common difference \(\displaystyle d\). \[\frac{a+10d}{a+17d}=\frac{2}{3} \] \[3(a+10d)=2(a+17d) \] \[a=4d \] Fifth and 21st terms: \[a_5=a+4d=8d,\qquad a_{21}=a+20d=24d \] \[a_5:a_{21}=8d:24d=1:3 \] Sums: \[S_5=\frac{5}{2}(2a+4d)=\frac{5}{2}(12d)=30d \] \[S_{21}=\frac{21}{2}(2a+20d)=\frac{21}{2}(28d)=294d \] \[S_5:S_{21}=30d:294d=5:49 \] Answer: \(\displaystyle a_5:a_{21}=1:3\); \(\displaystyle S_5:S_{21}=5:49\)

  7. Exercise 7

    Show that the sum of an AP whose first term is a\displaystyle a, the second term b\displaystyle b and the last term c\displaystyle c, is equal to (a+c)(b+c−2a)2(b−a)\frac{(a+c)(b+c-2 a)}{2(b-a)}

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    NCERT prints no numerical answer for this exercise, so this working has not been cross-checked against the book.

    Let common difference \(\displaystyle d=b-a\). \[c=a+(n-1)d=a+(n-1)(b-a) \] \[n-1=\frac{c-a}{b-a} \] \[n=\frac{(b-a)+(c-a)}{b-a}=\frac{b+c-2a}{b-a} \] \[S_n=\frac{n}{2}(a+c) \quad \text{(sum of }n\text{ terms)} \] \[S_n=\frac{(a+c)(b+c-2a)}{2(b-a)} \] Answer: \(\displaystyle S_n=\dfrac{(a+c)(b+c-2a)}{2(b-a)}\)

  8. Exercise 8

    Solve the equation −4+(−1)+2+…+x=437-4+(-1)+2+\ldots+x=437

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    NCERT’s answer
    $\displaystyle 50$
    AP: \(\displaystyle a=-4,\ d=3\). \[x=a_n=-4+3(n-1) \] \[S_n=\frac{n}{2}(a+x)=437 \] \[\frac{n}{2}(x-4)=437 \] Substitute \(\displaystyle x=3n-7\): \[n(3n-11)=874 \] \[3n^2-11n-874=0 \] \[n=\frac{11\pm\sqrt{121+10488}}{6}=\frac{11\pm103}{6} \] \[n=19 \quad (n>0) \] \[x=3(19)-7=50 \] Answer: \(\displaystyle x=50\)

  9. Exercise 9

    Jaspal Singh repays his total loan of Rs 118000\displaystyle 118000 by paying every month starting with the first instalment of Rs 1000. If he increases the instalment by Rs 100\displaystyle 100 every month, what amount will be paid by him in the 30th \displaystyle 30^{\text {th }} instalment? What amount of loan does he still have to pay after the 30th \displaystyle 30{ }^{\text {th }} instalment?

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    NCERT’s answer
    Rs $\displaystyle 3900$; Rs $\displaystyle 44500$
    AP: \(\displaystyle a=1000,\ d=100\). \[a_{30}=a+29d=1000+2900=3900 \] \[S_{30}=\frac{30}{2}(2a+29d)=15(2000+2900)=73500 \] \[\text{Remaining}=118000-73500=44500 \] Answer: 30th instalment Rs $\displaystyle 3900$; loan still to pay Rs $\displaystyle 44500$

  10. Exercise 10

    The students of a school decided to beautify the school on the Annual Day by fixing colourful flags on the straight passage of the school. They have 27\displaystyle 27 flags to be fixed at intervals of every 2\displaystyle 2 m. The flags are stored at the position of the middle most flag. Ruchi was given the responsibility of placing the flags. Ruchi kept her books where the flags were stored. She could carry only one flag at a time. How much distance did she cover in completing this job and returning back to collect her books? What is the maximum distance she travelled carrying a flag?

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    NCERT’s answer
    $\displaystyle 728$ m; $\displaystyle 26$ m.
    $\displaystyle 27$ flags, interval \(\displaystyle 2\text{ m}\); store at the \(\displaystyle 14^{\text{th}}\) (middle) flag.Distance from store to flag \(\displaystyle i\): \[d_i=2|i-14| \] One side (\(\displaystyle i=1,\dots,13\)): \[\sum d_i = 2+4+\cdots+26=2\cdot\frac{13\cdot14}{2}=182 \] Both sides ($\displaystyle 26$ flags): \[2\times182=364 \] Every flag needs a return trip to the store: \[\text{Total distance}=2\times364=728\text{ m} \] Farthest flag (\(\displaystyle i=1\) or \(\displaystyle i=27\)): \[d_{\max}=2\times13=26\text{ m} \] Answer: $\displaystyle 728$ m; $\displaystyle 26$ m