Sequence of series practice · Mixed difficulty

Option A ($\frac{1}{2}$) is correct because by expanding the logarithmic expression and identifying the coefficient of $x^{-4}$, we find it to be $\frac{1}{2}$.

Option B (-55) is correct because the sum of the products of the numbers $\pm 1, \pm 2, \pm 3, \pm 4, \pm 5$ taken two at a time is -55.

Option C ($1-\log _3 4$) is correct because by using the properties of logarithms and solving the resulting quadratic equation, we find $x = 1 - \log_3 4$.

Option D ($\frac{11}{41}$) is correct because by using the formula for the sum of an AP and substituting p=11 and q=41, we can find the ratio of a6/a21.

Option A ($\frac{\pi}{8}$) is correct because it is derived by recognizing the given series as a manipulation of the Leibniz formula for pi.

Option C ($\frac{1}{x}, \frac{1}{y}, \frac{1}{z}$ are in A.P.) is correct because it follows from the given conditions that a, b, c are in A.P., and a, b, c can be expressed in terms of x, y, z.

Option B (2abc) is correct because, given a is the AM of b and c, and G1 and G2 are the two geometric means, G1³ + G2³ simplifies to 2abc.

Option B (2475) is correct because it is the sum of all two-digit odd numbers, calculated using the arithmetic series formula.

Option A (x>3) is correct because, given the conditions of the increasing G.P., x must be greater than 3.

Option D (4) is correct because after simplifying the logarithmic expression and applying the properties of exponents, the result is 4.

Option C (4) is correct because by setting up equations based on the given conditions for the arithmetic series, we can solve for the first term.

Option D (r+s) is correct because, in an A.P., if the A.M. of two terms is equal, their indices must sum to the same value.

The correct answer is Zero (Option B) because if m times the mth term equals n times the nth term, then the (m+n)th term of the A.P. is zero.

Option C (n(n+1)(2n+7)/6) is correct because it is the simplified form of the summation after applying the formulas for the sum of first n natural numbers and the sum of squares.

Option C ($\frac{n(n+1)(2 n+1)}{6}$) is correct because it is the standard formula for the sum of the squares of the first n natural numbers.

Option A ($\frac{{16}}{{5}}$) is correct because the Harmonic Mean (HM) is calculated using the formula HM = $\frac{{2ab}}{{a+b}}$ and the given AM and GM values.

Option B ($\frac{n}{(1+x)[1+(n+1) x]}$) is correct because it is the sum to n terms of the given series, obtained by using partial fractions.

Option B (3/7) is correct because it is the 6th term of the Harmonic Progression, calculated using the properties of AP and HP.

Option B (4) is correct because the minimum value of the expression is found using AM >= GM inequality, resulting in a minimum value of 4.

Option C ($2,\;5,\;8$ and $26,\;5,\; - 16$) is correct because these numbers satisfy the conditions of being in AP and then in GP after adding 1, 4, and 19 respectively.

Option B ($\frac{\sqrt{5}-1}{2} < r < \frac{\sqrt{5}+1}{2}$) is correct because it satisfies the triangle inequality theorem for the sides of a triangle formed by three successive terms of a G.P.

Option C ($2 n a_{n+1}$) is correct because it is derived from the property that the sum of terms equidistant from the beginning and end of an AP is constant.

Option D (43,45,.......,89) is correct because it represents the consecutive odd integers that sum up to the given difference of squares, found by using the properties of arithmetic progressions and solving for the first and last terms.

Option D ($1 + \frac{{{{\left( {2x} \right)}^2}}}{{2.2!}} + \frac{{{{\left( {2x} \right)}^4}}}{{2.4!}} + ...\infty$) is correct because it represents the given expression in ascending powers of x after simplification using exponential series.

Option C ($3\mathrm{ }:\mathrm{ }4\mathrm{ }:\mathrm{ }5$) is correct because it represents the sides of a right-angled triangle in AP, satisfying the Pythagorean theorem.

Option C (13) is correct because the 13th term of the given GP is 2187.

Option A ($2,6,10,14$) is correct because these numbers are in AP, sum to 32, and satisfy the given ratio of products.

Option D (None of these) is incorrect. The correct answer is C (852/7) because it is the sum of the positive terms of the given arithmetic series.

Option D (75) is correct because simplifying the logarithmic equation leads to $\lambda = 2^{6/11}$, and expressing this in the form $\lambda = \sqrt[b]{a}$ gives $a = 2^6 = 64$ and $b = 11$, so $a + b = 75$.

Option C ($p{\left( {\frac{q}{p}} \right)^{m/2n}}$) is correct because by using the properties of geometric progressions, we can find the nth term.