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NCERT Exemplar · Class 10 Mathematics Arithmetic Progressions

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EXERCISE 5.2 1–8 (part 3 of 8)

  1. Exercise 1

    Which of the following form an AP? Justify your answer.
    (i)
    −1,−1,−1,−1,…\displaystyle -1,-1,-1,-1, \ldots
    (ii)
    0,2,0,2,…\displaystyle 0,2,0,2, \ldots
    (iii)
    1,1,2,2,3,3,…\displaystyle 1,1,2,2,3,3, \ldots
    (iv)
    11,22,33,…\displaystyle 11,22,33, \ldots
    (v)
    12,13,14,…\displaystyle \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \ldots
    (vi)
    2,22,23,24,…\displaystyle 2,2^2, 2^3, 2^4, \ldots
    (vii)
    3,12,27,48,…\displaystyle \sqrt{3}, \sqrt{12}, \sqrt{27}, \sqrt{48}, \ldots

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    NCERT’s answer
    (i), (iv) and (vii) form an AP as in each of these \(\displaystyle a_{k+1}-a_k\) is the same for different values of \(\displaystyle k\).
    Common difference must be equal between every consecutive pair. \[(i)\ a_2-a_1=0,\ a_3-a_2=0,\ a_4-a_3=0 \] \[(ii)\ a_2-a_1=2,\ a_3-a_2=-2 \] \[(iii)\ a_2-a_1=0,\ a_3-a_2=1 \] \[(iv)\ a_2-a_1=11,\ a_3-a_2=11 \] \[(v)\ a_2-a_1=-\frac16,\ a_3-a_2=-\frac1{12} \] \[(vi)\ a_2-a_1=2,\ a_3-a_2=4 \] \[(vii)\ \sqrt{12}=2\sqrt3,\ \sqrt{27}=3\sqrt3,\ \sqrt{48}=4\sqrt3;\quad a_2-a_1=a_3-a_2=a_4-a_3=\sqrt3 \] Answer: (i), (iv), (vii) form an AP (\(\displaystyle d=0,11,\sqrt3\)); (ii), (iii), (v), (vi) do not, since their consecutive differences are unequal.

  2. Exercise 2

    Justify whether it is true to say that −1,−32,−2,52,…\displaystyle -1,-\frac{3}{2},-2, \frac{5}{2}, \ldots forms an AP as a2−a1=a3−a2\displaystyle a_2-a_1=a_3-a_2.

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    NCERT’s answer
    False, as \(\displaystyle a_4-a_3 \neq a_3-a_2\).
    False. \[a_2-a_1=-\frac32-(-1)=-\frac12,\quad a_3-a_2=-2-\left(-\frac32\right)=-\frac12 \] \[a_4-a_3=\frac52-(-2)=\frac92\neq-\frac12 \] One matching pair does not make an AP; every consecutive difference must be equal, and here it is not.

  3. Exercise 3

    For the AP: -3\displaystyle 3, -7\displaystyle 7, -11\displaystyle 11, ..., can we find directly a30−a20\displaystyle a_{30}-a_{20} without actually finding a30\displaystyle a_{30} and a20\displaystyle a_{20}? Give reasons for your answer.

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    NCERT’s answer
    Yes, \(\displaystyle a_{30}-a_{20}=(30-20) d=10 d=-40\).
    \[a_n=a+(n-1)d,\qquad a_{30}-a_{20}=(a+29d)-(a+19d)=10d \] \[d=a_2-a_1=-7-(-3)=-4 \] \[a_{30}-a_{20}=10(-4)=-40 \] The difference of two terms of an AP depends only on \(\displaystyle d\) and the gap between their positions, not on the terms themselves. Answer: Yes; \(\displaystyle a_{30}-a_{20}=10d=-40\).

  4. Exercise 4

    Two APs have the same common difference. The first term of one AP is 2\displaystyle 2 and that of the other is 7. The difference between their 10th \displaystyle 10^{\text {th }} terms is the same as the difference between their 21st \displaystyle 21^{\text {st }} terms, which is the same as the difference between any two corresponding terms. Why?

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    NCERT’s answer
    The difference between any two corresponding terms of such APs is the same as the difference between their first terms.
    \[a_n=2+(n-1)d,\qquad b_n=7+(n-1)d \] \[b_n-a_n=\bigl[7+(n-1)d\bigr]-\bigl[2+(n-1)d\bigr]=7-2=5 \] The \(\displaystyle (n-1)d\) part is common to both APs and cancels for every \(\displaystyle n\), leaving only the gap between the first terms. Answer: The difference is always \(\displaystyle 5\) — the first terms' gap — whatever \(\displaystyle n\) or \(\displaystyle d\) is.

  5. Exercise 5

    Is 0\displaystyle 0 a term of the AP: 31\displaystyle 31, 28\displaystyle 28, 25\displaystyle 25, ...? Justify your answer.

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    NCERT’s answer
    No.
    \[a=31,\ d=28-31=-3,\qquad a_n=31+(n-1)(-3)=34-3n \] \[34-3n=0\ \Rightarrow\ n=\frac{34}{3} \] \(\displaystyle n\) is not a natural number, so \(\displaystyle 0\) never occurs as a term. Answer: No; \(\displaystyle n=\dfrac{34}{3}\) is not a positive integer.

  6. Exercise 6

    The taxi fare after each km, when the fare is Rs 15\displaystyle 15 for the first km and Rs 8\displaystyle 8 for each additional km, does not form an AP as the total fare (in Rs) after each km is 15,8,8,8,…15,8,8,8, \ldots Is the statement true? Give reasons.

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    NCERT’s answer
    No, as the total fare (in Rs) after each km is $\displaystyle 15$, $\displaystyle 23$, $\displaystyle 31$, $\displaystyle 39$, ---
    False. \[\text{total fare after } n\text{ km}=15+8(n-1) \] \[15,\ 15+8,\ 15+2(8),\ 15+3(8),\ldots=15,23,31,39,\ldots \] The quoted list \(\displaystyle 15,8,8,8,\ldots\) is the per-km charge, not the running total; the actual total fare is an AP with common difference \(\displaystyle 8\).

  7. Exercise 7

    In which of the following situations, do the lists of numbers involved form an AP? Give reasons for your answers.
    (i)
    The fee charged from a student every month by a school for the whole session, when the monthly fee is Rs 400.
    (ii)
    The fee charged every month by a school from Classes I to XII, when the monthly fee for Class I is Rs 250\displaystyle 250, and it increases by Rs 50\displaystyle 50 for the next higher class.
    (iii)
    The amount of money in the account of Varun at the end of every year when Rs 1000\displaystyle 1000 is deposited at simple interest of 10%\displaystyle 10 \% per annum.
    (iv)
    The number of bacteria in a certain food item after each second, when they double in every second.

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    NCERT’s answer
    (i), (ii) and (iii) form an AP as in the list of numbers formed every succeeding term is obtained by adding a fixed number.
    \[(i)\ 400,400,400,\ldots;\quad a_2-a_1=0 \] \[(ii)\ 250,300,350,\ldots;\quad a_2-a_1=a_3-a_2=50 \] \[(iii)\ \text{amount after } n\text{ yr}=1000+1000\times\frac{10}{100}\times n=1000+100n;\quad \text{difference}=100 \] \[(iv)\ N,2N,4N,8N,\ldots;\quad a_2-a_1=N,\ a_3-a_2=2N \] Answer: (i), (ii), (iii) form an AP; (iv) does not, since \(\displaystyle a_2-a_1\neq a_3-a_2\).

  8. Exercise 8

    Justify whether it is true to say that the following are the nth \displaystyle n^{\text {th }} terms of an AP.
    (i)
    2n−3\displaystyle 2 n-3
    (ii)
    3n2+5\displaystyle 3 n^2+5
    (iii)
    1+n+n2\displaystyle 1+n+n^2

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    NCERT’s answer
    (i)
    Yes
    (ii)
    No
    (iii)
    No
    \[(i)\ a_n=2n-3;\quad a_{n+1}-a_n=2(n+1)-3-(2n-3)=2 \] \[(ii)\ a_n=3n^2+5;\quad a_{n+1}-a_n=3(n+1)^2+5-(3n^2+5)=6n+3 \] \[(iii)\ a_n=1+n+n^2;\quad a_{n+1}-a_n=\bigl[1+(n+1)+(n+1)^2\bigr]-\bigl[1+n+n^2\bigr]=2n+2 \] Only (i) has a difference free of \(\displaystyle n\); (ii) and (iii) grow with \(\displaystyle n\), so they are not AP terms. Answer: (i) is an AP (\(\displaystyle d=2\)); (ii) and (iii) are not.