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End Practice
Sections
Progression (Level - 2)
Progression (Level - 2)
Q1–10 of 50
1

If a times the ath term of an A.P. is equal to b times the bth term, find the (a + b)th term.

Correct Answer: A. 0
Explanation:


  • Statement: $a \cdot T_a = b \cdot T_b$, where $T_n = A + (n-1)d$. Find $T_{a+b}$.

  • Calculation:

    $$a[A + (a-1)d] = b[A + (b-1)d]$$
    $$aA + a(a-1)d = bA + b(b-1)d$$
    $$(a-b)A + [a^2 - a - b^2 + b]d = 0$$
    $$(a-b)A + [(a^2 - b^2) - (a - b)]d = 0$$
    $$(a-b) [A + (a + b - 1)d] = 0$$

    Since $a \neq b$, we divide by $(a-b)$:

    $$A + (a + b - 1)d = 0 \implies T_{a+b} = 0$$

  • Correct Answer: (a) 0

    • 2

    A number 28 is divided into four parts that are in AP such that the product of the first and fourth is to the product of the second and third is 5 : 6. Find the smallest part.

    Correct Answer: B. 4
    Explanation:


    $$\text{Sum} = 4a = 28 \implies a = 7$$

    Given ratio:

    $$\frac{(a-3d)(a+3d)}{(a-d)(a+d)} = \frac{5}{6} \implies \frac{a^2 - 9d^2}{a^2 - d^2} = \frac{5}{6}$$

    Substitute $a = 7 \implies a^2 = 49$:

    $$6(49 - 9d^2) = 5(49 - d^2)$$
    $$294 - 54d^2 = 245 - 5d^2$$
    $$49d^2 = 49 \implies d^2 = 1 \implies d = 1$$
    $$\text{Smallest part} = a - 3d = 7 - 3(1) = 4$$
    3

    Find the value of the expression: 1 – 3 + 5 – 7…. to 100 terms.

    Correct Answer: B. –100
    Explanation:


  • Expression: $1 - 3 + 5 - 7 + \dots \text{ to } 100 \text{ terms}$

  • Calculation: Group into 50 pairs of 2 terms each:

    $$(1 - 3) + (5 - 7) + (9 - 11) + \dots \text{ (50 pairs)}$$
    $$= (-2) + (-2) + (-2) + \dots \text{ (50 times)}$$
    $$= 50 \times (-2) = -100$$
    • 4

    If a clock strikes once at 12 A.M., twice at 1 A.M., thrice at 2 A.M. and so on, how many times will the clock be struck in the course of 3 days? (Assume a 24 hour clock)

    Correct Answer: B. 828
    Explanation:


    Standard 1 to 12 clock cycle sum per 12 hours:

    $$\text{Sum per 12 hrs} = 1 + 2 + 3 + \dots + 12 = \frac{12 \times 13}{2} = 78$$

    In 1 day (24 hours = two 12-hour cycles):

    $$\text{Strikes per day} = 78 \times 2 = 156$$

    For 3 days:

    $$156 \times 3 = 468$$

    (If assuming a true 1 to 24 hour clock without reset):

    $$\text{Sum per day} = \frac{24 \times 25}{2} = 300 \implies 300 \times 3 = 900$$
    5

    What will be the maximum sum of 54, 52, 50, … ?

    Correct Answer: C. 756
    Explanation:


  • Series: $54, 52, 50, \dots$ (AP with $a = 54$, $d = -2$)

  • Calculation: Maximum sum occurs when we include only all non-negative terms.

    $$T_n \ge 0 \implies 54 + (n-1)(-2) \ge 0 \implies 56 \ge 2n \implies n \le 28$$

    Last positive term is $T_{27} = 2$, $T_{28} = 0$.

    $$S_{27} = \frac{27}{2} [54 + 2] = \frac{27}{2} \times 56 = 27 \times 28 = 756$$
  • 6

    Find the sum of the integers between 100 and 300 that are multiples of 7.

    Correct Answer: B. 5586
    Explanation:


  • Multiples of 7 between 100 and 300:

    • First multiple $> 100$: $105 = 7 \times 15$

    • Last multiple $< 300$: $294 = 7 \times 42$

  • Number of terms ($n$): $42 - 15 + 1 = 28$

  • Sum:

    $$S_n = \frac{n}{2} (\text{first} + \text{last}) = \frac{28}{2} (105 + 294) = 14 \times 399 = 5586$$

  • Correct Answer: (b) 5586

    • 7

    If x > 1, y > 1, z > 1 are in G.P., then 1/(1 + log x), 1/(1 + log y), 1/(1 + log z) are in

    Correct Answer: B. H.P
    Explanation:


  • Given: $x, y, z$ in G.P. $\implies y^2 = x z \implies 2 \log y = \log x + \log z$

  • Analysis:

    Therefore, $\log x, \log y, \log z$ are in A.P.

    Adding 1 to each term preserves the A.P.: $(1 + \log x), (1 + \log y), (1 + \log z)$ are in A.P.

    Reciprocals of terms in A.P. form a Harmonic Progression (H.P.).

  • Correct Answer: (b) H.P.

    • 8

    Find the sum of all odd numbers lying between 1000 and 2000.

    Correct Answer: A. 7,50,000
    Explanation:


  • Odd numbers lying strictly between 1000 and 2000:

    • First term $a = 1001$

    • Last term $l = 1999$

    • Common difference $d = 2$

  • Number of terms ($n$):

    $$1999 = 1001 + (n-1)2 \implies 998 = 2(n-1) \implies n = 500$$

  • Sum:

    $$S = \frac{500}{2} (1001 + 1999) = 250 \times 3000 = 7,50,000$$

  • Correct Answer: (a) 7,50,000

    • 9

    Find the sum of all integers of 3 digits that are divisible by 11.

    Correct Answer: C. 44,550
    Explanation:


  • 3-digit integers divisible by 11:

    • First term $a = 110$ ($11 \times 10$)

    • Last term $l = 990$ ($11 \times 90$)

  • Number of terms ($n$):

    $$n = 90 - 10 + 1 = 81$$

  • Sum:

    $$S = \frac{81}{2} (110 + 990) = \frac{81}{2} \times 1100 = 81 \times 550 = 44,550$$

  • Correct Answer: (c) 44,550

    • 10

    The first and the last terms of an A.P. are 113 and 253. If there are six terms in this sequence, find the sum of sequence.

    Correct Answer: C. 1098
    Explanation:


  • Given: $a = 113$, $l = 253$, $n = 6$.

  • Sum of AP:

    $$S_n = \frac{n}{2} (a + l) = \frac{6}{2} (113 + 253) = 3 \times 366 = 1098$$

  • Correct Answer: (c) 1098

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