SolveIt is under development
SolveItNCERT · CBSE · NEET

NCERT Exemplar · Class 10 Mathematics Arithmetic Progressions

71 questions · 71 still being checked

EXERCISE 5.3 1–10 (part 4 of 8)

  1. Exercise 1

    Match the APs given in column A with suitable common differences given in column B.
    Column AColumn B
    (A1)2,−2,−6,−10,…\displaystyle \left(\mathrm{A}_1\right) \quad 2,-2,-6,-10, \ldots(B1)23\displaystyle \left(\mathrm{B}_1\right) \quad \frac{2}{3}
    (A2)a=−18,n=10,an=0\displaystyle \left(\mathrm{A}_2\right) \quad a=-18, n=10, a_n=0(B2)−5\displaystyle \left(\mathrm{B}_2\right) \quad-5
    (A3)a=0,a10=6\displaystyle \left(\mathrm{A}_3\right) \quad a=0, a_{10}=6(B3)4\displaystyle \left(\mathrm{B}_3\right) \quad 4
    (A4)a2=13,a4=3\displaystyle \left(\mathrm{A}_4\right) \quad a_2=13, a_4=3(B4)−4\displaystyle \left(\mathrm{B}_4\right) \quad-4
    (B5)2\displaystyle \left(\mathrm{B}_5\right) \quad 2
    (B6)12\displaystyle \left(\mathrm{B}_6\right) \quad \frac{1}{2}
    (B7)5\displaystyle \left(\mathrm{B}_7\right) \quad 5

    Matches the book, not yet reviewed

    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    (\(\displaystyle A_{1}\)) → (\(\displaystyle B_{4}\)) (\(\displaystyle A_{2}\)) → (\(\displaystyle B_{5}\)) (\(\displaystyle A_{3}\)) → (\(\displaystyle B_{1}\)) (\(\displaystyle A_{4}\)) → (\(\displaystyle B_{2}\))
    \[A_1:\ d = -2 - 2 = -4 \] \[A_2:\ 0 = -18 + 9d \Rightarrow d = 2 \] \[A_3:\ 6 = 0 + 9d \Rightarrow d = \frac{2}{3} \] \[A_4:\ d = \frac{3-13}{2} = -5 \] Answer: \(\displaystyle A_1\text{-}B_4,\ A_2\text{-}B_5,\ A_3\text{-}B_1,\ A_4\text{-}B_2\)

  2. Exercise 2

    Verify that each of the following is an AP, and then write its next three terms.
    (i)
    0,14,12,34,…\displaystyle 0, \frac{1}{4}, \frac{1}{2}, \frac{3}{4}, \ldots
    (ii)
    5,143,133,4,…\displaystyle 5, \frac{14}{3}, \frac{13}{3}, 4, \ldots
    (iii)
    3,23,33,…\displaystyle \sqrt{3}, 2 \sqrt{3}, 3 \sqrt{3}, \ldots
    (iv)
    a+b,(a+1)+b,(a+1)+(b+1),…\displaystyle a+b,(a+1)+b,(a+1)+(b+1), \ldots
    (v)
    a,2a+1,3a+2,4a+3,…\displaystyle a, 2 a+1,3 a+2,4 a+3, \ldots

    Matches the book, not yet reviewed

    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    (i)
    \(\displaystyle 1, \frac{5}{4}, \frac{3}{2}\)
    (ii)
    \(\displaystyle \frac{11}{3}, \frac{10}{3}, 3\)
    (iii)
    \(\displaystyle 4 \sqrt{3}, 5 \sqrt{3}, 6 \sqrt{3}\)
    (iv)
    \(\displaystyle (a+2)+(b+1),(a+2)+(b+2),(a+3)+(b+2)\)
    (v)
    \(\displaystyle 5 a+4,6 a+5,7 a+6\)
    (i)
    \[\tfrac14-0=\tfrac14,\quad \tfrac12-\tfrac14=\tfrac14,\quad \tfrac34-\tfrac12=\tfrac14 \]
    \[d=\tfrac14\ \text{constant} \Rightarrow \text{AP} \]
    Next three: \[1,\ \tfrac54,\ \tfrac32 \]
    (ii)
    \[\tfrac{14}{3}-5=-\tfrac13,\quad \tfrac{13}{3}-\tfrac{14}{3}=-\tfrac13,\quad 4-\tfrac{13}{3}=-\tfrac13 \]
    \[d=-\tfrac13 \Rightarrow \text{AP} \]
    Next three: \[\tfrac{11}{3},\ \tfrac{10}{3},\ 3 \]
    (iii)
    \[2\sqrt3-\sqrt3=\sqrt3,\quad 3\sqrt3-2\sqrt3=\sqrt3 \]
    \[d=\sqrt3 \Rightarrow \text{AP} \]
    Next three: \[4\sqrt3,\ 5\sqrt3,\ 6\sqrt3 \]
    (iv)
    \[[(a{+}1)+b]-(a+b)=1,\quad [(a{+}1)+(b{+}1)]-[(a{+}1)+b]=1 \]
    \[d=1 \Rightarrow \text{AP} \]
    Next three: \[(a{+}2)+(b{+}1),\ \ (a{+}2)+(b{+}2),\ \ (a{+}3)+(b{+}2) \]
    (v)
    \[(2a{+}1)-a=a{+}1,\quad (3a{+}2)-(2a{+}1)=a{+}1,\quad (4a{+}3)-(3a{+}2)=a{+}1 \]
    \[d=a+1 \Rightarrow \text{AP} \]
    Next three: \[5a{+}4,\ 6a{+}5,\ 7a{+}6 \]
    Answer: (i) \(\displaystyle 1,\ \tfrac54,\ \tfrac32\) (ii) \(\displaystyle \tfrac{11}{3},\ \tfrac{10}{3},\ 3\) (iii) \(\displaystyle 4\sqrt3,\ 5\sqrt3,\ 6\sqrt3\) (iv) \(\displaystyle (a{+}2)+(b{+}1),\ (a{+}2)+(b{+}2),\ (a{+}3)+(b{+}2)\) (v) \(\displaystyle 5a{+}4,\ 6a{+}5,\ 7a{+}6\)

  3. Exercise 3

    Write the first three terms of the APs when a\displaystyle a and d\displaystyle d are as given below:
    (i)
    a=12,d=−16\displaystyle a=\frac{1}{2}, d=-\frac{1}{6}
    (ii)
    a=−5,d=−3\displaystyle a=-5, d=-3
    (iii)
    a=2,d=12\displaystyle a=\sqrt{2}, \quad d=\frac{1}{\sqrt{2}}

    Matches the book, not yet reviewed

    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    (i)
    \(\displaystyle \frac{1}{2}, \frac{1}{3}, \frac{1}{6}\)
    (ii)
    -$\displaystyle 5$, -$\displaystyle 8$, -$\displaystyle 11$
    (iii)
    \(\displaystyle \sqrt{2}, \frac{3}{\sqrt{2}}, \frac{4}{\sqrt{2}}\)
    (i)
    \[a_1=\tfrac12,\quad a_2=\tfrac12-\tfrac16=\tfrac13,\quad a_3=\tfrac13-\tfrac16=\tfrac16 \]
    (ii)
    \[a_1=-5,\quad a_2=-5-3=-8,\quad a_3=-8-3=-11 \]
    (iii)
    \[a_1=\sqrt2,\quad a_2=\sqrt2+\tfrac{1}{\sqrt2}=\tfrac{3}{\sqrt2},\quad a_3=\tfrac{3}{\sqrt2}+\tfrac{1}{\sqrt2}=\tfrac{4}{\sqrt2} \]
    Answer: (i) \(\displaystyle \tfrac12,\ \tfrac13,\ \tfrac16\) (ii) \(\displaystyle -5,\ -8,\ -11\) (iii) \(\displaystyle \sqrt2,\ \tfrac{3}{\sqrt2},\ \tfrac{4}{\sqrt2}\)

  4. Exercise 4

    Find a,b\displaystyle a, b and c\displaystyle c such that the following numbers are in AP: a,7,b,23,c\displaystyle a, 7, b, 23, c.

    Matches the book, not yet reviewed

    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    \(\displaystyle a=-1, b=15, c=31\)
    \[2d = 23-7 = 16 \Rightarrow d = 8 \] \[a = 7-d = -1,\quad b = 7+d = 15,\quad c = 23+d = 31 \] Answer: \(\displaystyle a=-1,\ b=15,\ c=31\)

  5. Exercise 5

    Determine the AP whose fifth term is 19\displaystyle 19 and the difference of the eighth term from the thirteenth term is 20.

    Matches the book, not yet reviewed

    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    $\displaystyle 3$, $\displaystyle 7$, $\displaystyle 11$, $\displaystyle 15$, ---
    \[a_5 = a+4d = 19 \] \[a_{13}-a_8 = 5d = 20 \Rightarrow d = 4 \] \[a = 19-4(4) = 3 \] Answer: AP: \(\displaystyle 3,\ 7,\ 11,\ 15,\ 19,\ \ldots\)

  6. Exercise 6

    The 26th ,11th \displaystyle 26^{\text {th }}, 11^{\text {th }} and the last term of an AP are 0\displaystyle 0, 3\displaystyle 3 and −15\displaystyle -\frac{1}{5}, respectively. Find the common difference and the number of terms.

    Matches the book, not yet reviewed

    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    \(\displaystyle d=-\frac{1}{5}, n=27\)
    \[a_{26}=a+25d=0,\qquad a_{11}=a+10d=3 \] \[15d=-3 \Rightarrow d=-\tfrac15 \] \[a=3-10d=3+2=5 \] \[a+(n-1)d=-\tfrac15 \] \[5-\tfrac{n-1}{5}=-\tfrac15 \Rightarrow n-1=26 \Rightarrow n=27 \] Answer: \(\displaystyle d=-\tfrac15,\ n=27\)

  7. Exercise 7

    The sum of the 5th \displaystyle 5^{\text {th }} and the 7th \displaystyle 7^{\text {th }} terms of an AP is 52\displaystyle 52 and the 10th \displaystyle 10^{\text {th }} term is 46. Find the AP.

    Matches the book, not yet reviewed

    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    $\displaystyle 1$, $\displaystyle 6$, $\displaystyle 11$, $\displaystyle 16$, ---
    \[a_5+a_7=2a+10d=52 \Rightarrow a+5d=26 \] \[a_{10}=a+9d=46 \] \[4d=20 \Rightarrow d=5 \] \[a=26-5(5)=1 \] Answer: AP: \(\displaystyle 1,\ 6,\ 11,\ 16,\ 21,\ \ldots\)

  8. Exercise 8

    Find the 20th \displaystyle 20{ }^{\text {th }} term of the AP whose 7th \displaystyle 7^{\text {th }} term is 24\displaystyle 24 less than the 11th \displaystyle 11^{\text {th }} term, first term being 12.

    Matches the book, not yet reviewed

    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    $\displaystyle 126$
    Given \(\displaystyle a=12\); the condition is \(\displaystyle a_{11}-a_7=24\). \[a_{11}-a_7 = (a+10d)-(a+6d) = 4d \] \[4d = 24 \implies d = 6 \] \[a_{20} = a+19d = 12+19(6) = 126 \]Answer: \(\displaystyle a_{20}=126\).

  9. Exercise 9

    If the 9th \displaystyle 9^{\text {th }} term of an AP is zero, prove that its 29th \displaystyle 29^{\text {th }} term is twice its 19th \displaystyle 19^{\text {th }} term.

    Not cross-checked

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

    Given \(\displaystyle a_9=0\). \[a_9 = a+8d = 0 \implies a = -8d \] \[a_{29} = a+28d = -8d+28d = 20d \] \[a_{19} = a+18d = -8d+18d = 10d \] \[a_{29} = 20d = 2(10d) = 2a_{19} \]Answer: \(\displaystyle a_{29}=2a_{19}\).

  10. Exercise 10

    Find whether 55\displaystyle 55 is a term of the AP: 7\displaystyle 7, 10\displaystyle 10, 13\displaystyle 13,--- or not. If yes, find which term it is.

    Matches the book, not yet reviewed

    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    Yes, \(\displaystyle 17{ }^{\text {th }}\) term.
    AP: \(\displaystyle a=7,\ d=3\). \[a_n = a+(n-1)d = 7+3(n-1) \] \[55 = 7+3(n-1) \implies n-1 = 16 \implies n = 17 \]Answer: Yes; $\displaystyle 55$ is the \(\displaystyle 17^{\text{th}}\) term.