Q.100 stones are placed on a straight road at intervals of 5 m apart. A runner has to start from a fixed point 5 m from the first stone in the same straight line, pick up the stones and bring them back to the fixed point one by one. How many kilometer has he to run altogether ?

Showing the 12 most recent of 17 on this concept.

1. CBSE 2025Set ANNUAL1 markMCQ
   Q.The common difference of the A.P. $8, 15, 22, \ldots$ is

   (a) 8

   (b) 15

   (c) 22

   (d) 7

   ›Reveal solutionSolution

   The common difference $d = 7$.

   For an A.P., $d = a_2 - a_1$ (equivalently $a_3-a_2$, etc.).

   …

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2. CBSE 2025Set ANNUAL1 markMCQ
   Q.The arithmetic mean of two numbers 12 and 15 is

   (a) 13

   (b) 13.5

   (c) 14.5

   (d) 14

   ›Reveal solutionSolution

   The arithmetic mean of 12 and 15 is $13.5$.

   For two numbers $a$ and $b$, the arithmetic mean is $\dfrac{a+b}{2}$.

   …

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3. CBSE 2025Set ANNUAL1 markMCQ
   Q.The 15th term of the A.P. $5, 13, 21, 29, \ldots$ is

   (a) 113

   (b) 115

   (c) 116

   (d) 117

   ›Reveal solutionSolution

   The 15th term is $117$.

   For the A.P. $5, 13, 21, 29, \ldots$: first term $a=5$, common difference $d=13-5=8$.

   …

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4. CBSE 2025Set ANNUAL1 markMCQ
   Q.Which term of the A.P. $9, 14, 19, 24, 29, \ldots$ is $379$?

   (a) 80th

   (b) 78th

   (c) 75th

   (d) 60th

   ›Reveal solutionSolution

   Using $a_n = a + (n-1)d$ with $a=9$, $d=5$, and $a_n=379$ gives $n=75$.

   The A.P. is $9, 14, 19, 24, 29, \ldots$ with first term $a=9$ and common difference $d=14-9=5$.

   $a_n = a + (n-1)d$ …

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5. CBSE 2025Set ANNUAL1 markMCQ
   Q.What will be the sum of all natural numbers between 100 and 1000 which are divisible by 5?

   (a) 94850

   (b) 98450

   (c) 99450

   (d) 94950

   ›Reveal solutionSolution

   The multiples of 5 strictly between 100 and 1000 form an A.P. from 105 to 995 with common difference 5; summing this A.P. gives $98450$.

   "Between 100 and 1000" is taken to exclude the endpoints themselves, so the sequence of multiples of 5 runs from $105$ to $995$:

   $105, 110, 115, \ldots, 995$

   Number of terms: …

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6. CBSE 2024Set ANNUAL1 markMCQ
   Q.Find the sum of 17 terms of the A. P. 5, 9, 13, 17, ........:

   (a) 629

   (b) 529

   (c) 615

   (d) 0

   ›Reveal solutionSolution

   $S_{17}=\frac{17}{2}(10+64)=629$.

   AP: $5,9,13,17,\dots$ has $a=5$, $d=4$.

   $S_n = \dfrac n2[2a+(n-1)d]$

   …

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7. CBSE 2023Set ANNUAL1 markMCQ
   Q.The 10th term of a sequence whose 7th and 12th terms are 34 and 64 respectively is

   (a) 42

   (b) 52

   (c) 63

   (d) 36

   ›Reveal solutionSolution

   Treating this as an arithmetic sequence, the common difference is 6, giving the 10th term as 52.

   Let the sequence be arithmetic with first term $a$ and common difference $d$. We're told:

   $a_7=34, \qquad a_{12}=64$

   …

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8. CBSE 2023Set ANNUAL1 markMCQ
   Q.The sum of odd integers from 1 to 100 is:

   (a) 2525

   (b) 5050

   (c) 2500

   (d) None of these

   ›Reveal solutionSolution

   The odd numbers from 1 to 99 form an AP with 50 terms; their sum is $n^2$.

   The odd integers between 1 and 100 are $1, 3, 5, \dots, 99$ — an AP with first term $a=1$, common difference $d=2$. The last term is 99, so the number of terms is:

   $n = \frac{99-1}{2}+1 = 50$ …

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9. CBSE 2023Set ANNUAL1 markMCQ
   Q.The sum of $n$ A.M.'s between $a$ and $b$ is equal to

   (a) $\frac{n}{2}(a+b)$

   (b) $\frac{n+1}{2}(a+b)$

   (c) $n + 1$

   (d) $\frac{n+1}{2}$

   ›Reveal solutionSolution

   Sum of $n$ AMs $=\frac{n}{2}(a+b)$; option (a).

   Inserting $n$ arithmetic means between $a$ and $b$ gives an AP with $n+2$ terms. The $n$ means themselves have the same average as $a$ and $b$, namely $\frac{a+b}{2}$ (NCERT Class 11 Sequences a …

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10. CBSE 2022Set ANNUAL1 mark
    Q.If nth term of an A.P. is 5n + 1, then its common difference is ............ .

    ›Reveal solutionSolution

    For $a_n = 5n + 1$, consecutive terms differ by the coefficient of $n$, which is $5$.

    Given $a_n = 5n + 1$.

    …

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11. CBSE 2022Set ANNUAL1 mark
    Q.If 5th term of an A.P. is 10 and 10th term is 5, then its 15th term is ............ .

    ›Reveal solutionSolution

    Solving for $a$ and $d$ from the given terms gives $a = 14$, $d = -1$, so the 15th term is $0$.

    Let the first term be $a$ and common difference $d$. The $n$th term is $a_n = a + (n-1)d$.

    5th term: $a + 4d = 10$

    10th term: $a + 9d = 5$

    Subtracting: $(a+9d) - (a+4d) = 5 - 10 \Rightarrow 5d = -5 \Rightarrow d = -1$

    …

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12. CBSE 2022Set ANNUAL1 markMCQ
    Q.Fill in the blank with the correct option: the value of the 5th term is $\_\_\_\_$ in the A.P. $\sqrt{2}, \sqrt{8}, \sqrt{18}, \ldots$

    (a) $5\sqrt{2}$

    (b) 8

    (c) $7\sqrt{2}$

    ›Reveal solutionSolution

    Rewriting each term as a multiple of $\sqrt2$ reveals an A.P., whose 5th term is $5\sqrt2$.

    $\sqrt2 = \sqrt2$

    $\sqrt8 = \sqrt{4\times2} = 2\sqrt2$

    $\sqrt{18} = \sqrt{9\times2} = 3\sqrt2$

    …

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