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NCERT Exemplar · Class 10 Mathematics Arithmetic Progressions

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EXERCISE 5.3 31–35 (part 7 of 8)

  1. Exercise 31

    Find the sum of first seven numbers which are multiples of 2\displaystyle 2 as well as of 9. [Hint: Take the LCM of 2\displaystyle 2 and 9\displaystyle 9]

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    NCERT’s answer
    $\displaystyle 504$
    Numbers that are multiples of both $\displaystyle 2$ and $\displaystyle 9$ are multiples of their LCM. \[\text{LCM}(2,9) = 18 \] \[S_7 = \frac{7}{2}\big[2(18) + 6(18)\big] = \frac{7}{2}(144) \] \[S_7 = 504 \] Answer: \(\displaystyle 504\)

  2. Exercise 32

    How many terms of the AP: −15,−13,−11\displaystyle -15,-13,-11,--- are needed to make the sum −55\displaystyle -55? Explain the reason for double answer.

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    \(\displaystyle n=5,11\)
    \[S_n = \frac{n}{2}\big[2(-15) + (n-1)2\big] = n(n-16) \] \[n(n-16) = -55 \] \[n^2 - 16n + 55 = 0 \] \[n = \frac{16 \pm \sqrt{256-220}}{2} = \frac{16 \pm 6}{2} \] \[n = 5\ \text{or}\ n = 11 \] Terms \(\displaystyle a_6\) to \(\displaystyle a_{11}\) are \(\displaystyle -5,-3,-1,1,3,5\), which sum to $\displaystyle 0$, so \(\displaystyle S_{11}=S_5\); both counts give the same total. Answer: \(\displaystyle n = 5\) or \(\displaystyle n = 11\)

  3. Exercise 33

    The sum of the first n\displaystyle n terms of an AP whose first term is 8\displaystyle 8 and the common difference is 20\displaystyle 20 is equal to the sum of first 2n\displaystyle 2 n terms of another AP whose first term is −30\displaystyle -30 and the common difference is 8. Find n\displaystyle n.

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    $\displaystyle 11$
    \[S_n = \frac{n}{2}\big[2(8)+(n-1)20\big] = n(10n-2) \] \[S'_{2n} = \frac{2n}{2}\big[2(-30)+(2n-1)8\big] = n(16n-68) \] \[n(10n-2) = n(16n-68) \] \[10n-2 = 16n-68 \quad (n\neq 0) \] \[6n = 66 \] \[n = 11 \] Answer: \(\displaystyle n = 11\)

  4. Exercise 34

    Kanika was given her pocket money on Jan 1st \displaystyle 1^{\text {st }}, 2008. She puts Re 1\displaystyle 1 on Day 1\displaystyle 1, Rs 2\displaystyle 2 on Day 2\displaystyle 2, Rs 3\displaystyle 3 on Day 3\displaystyle 3, and continued doing so till the end of the month, from this money into her piggy bank. She also spent Rs 204\displaystyle 204 of her pocket money, and found that at the end of the month she still had Rs 100\displaystyle 100 with her. How much was her pocket money for the month?

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    Rs $\displaystyle 800$
    January has $\displaystyle 31$ days. \[\text{Saved} = \sum_{k=1}^{31} k = \frac{31}{2}(1+31) = 496 \] Pocket money splits into savings, spending and cash left in hand. \[\text{Pocket money} = 496 + 204 + 100 = 800 \] Answer: Rs $\displaystyle 800$

  5. Exercise 35

    Yasmeen saves Rs 32\displaystyle 32 during the first month, Rs 36\displaystyle 36 in the second month and Rs 40\displaystyle 40 in the third month. If she continues to save in this manner, in how many months will she save Rs 2000\displaystyle 2000?

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    $\displaystyle 25$ months.
    \[S_n = \frac{n}{2}\big[2(32)+(n-1)4\big] = n(2n+30) \] \[n(2n+30) = 2000 \] \[n^2 + 15n - 1000 = 0 \] \[n = \frac{-15+\sqrt{225+4000}}{2} = \frac{-15+65}{2} \] \[n = 25 \] Answer: $\displaystyle 25$ months