Let $a_i = i + \frac{1}{i}$ for $i = 1, 2, \ldots, 20$. Let $p = \frac{1}{20} \sum_{i=1}^{20} a_i$ and $q = \frac{1}{20} \sum_{i=1}^{20} \frac{1}{a_i}$. Then,

  • A
    $q \in \left(0, \frac{22-p}{21}\right)$
  • B
    $q \in \left[\frac{22-p}{21}, \frac{2(22-p)}{21}\right)$
  • C
    $q \in \left[\frac{2(22-p)}{21}, \frac{22-p}{7}\right)$
  • D
    $q \in \left[\frac{22-p}{7}, \frac{4(22-p)}{21}\right)$

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Similar Questions

Let $3, 7, 11, 15, \ldots, 403$ and $2, 5, 8, 11, \ldots, 404$ be two arithmetic progressions. Then the sum of the common terms in them is equal to:

Let $a, b \in \mathbb{R}$ be such that $a, a + 2b, 2a + b$ are in $A.P.$ and $(b + 1)^2, ab + 5, (a + 1)^2$ are in $G.P.$ Then $(a + b)$ equals:

If $|x| < 1, |y| < 1$ and $x \neq y,$ then the sum to infinity of the following series $(x+y)+(x^{2}+xy+y^{2})+(x^{3}+x^{2}y+xy^{2}+y^{3})+\ldots$ is:

Let $V_r$ denote the sum of the first $r$ terms of an arithmetic progression $(A.P.)$ whose first term is $r$ and the common difference is $(2r-1)$. Let $T_r = V_{r+1} - V_r - 2$ and $Q_r = T_{r+1} - T_r$ for $r = 1, 2, \ldots$
$1.$ The sum $V_1 + V_2 + \ldots + V_n$ is
$(A)$ $\frac{1}{12} n(n+1)(3n^2-n+1)$
$(B)$ $\frac{1}{12} n(n+1)(3n^2+n+2)$
$(C)$ $\frac{1}{2} n(2n^2-n+1)$
$(D)$ $\frac{1}{3}(2n^3-2n+3)$
$2.$ $T_r$ is always
$(A)$ an odd number
$(B)$ an even number
$(C)$ a prime number
$(D)$ a composite number
$3.$ Which one of the following is a correct statement?
$(A)$ $Q_1, Q_2, Q_3, \ldots$ are in $A.P.$ with common difference $5$
$(B)$ $Q_1, Q_2, Q_3, \ldots$ are in $A.P.$ with common difference $6$
$(C)$ $Q_1, Q_2, Q_3, \ldots$ are in $A.P.$ with common difference $11$
$(D)$ $Q_1 = Q_2 = Q_3 = \ldots$

If the first three terms of the sequence $\frac{1}{16}, a, b, \frac{1}{6}$ are in a geometric progression and the last three terms are in a harmonic progression,then the values of $a$ and $b$ are:

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