Sequence of series practice · Medium

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

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 ($\frac{{e - 1}}{{e + 1}}$) is correct because the expression can be simplified using the definitions of hyperbolic sine and cosine in terms of exponentials.

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 A ($\frac{1}{3}$) is correct because it is the common ratio that satisfies the given condition where each term is twice the sum of the succeeding terms.

Option D (Never less than $n$) is correct because, by the AM-GM inequality, the arithmetic mean of $n$ positive numbers is always greater than or equal to their geometric mean, and in this case, the geometric mean is 1.

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 C ($q^2=pr$) is correct because in a GP, the square of any term is equal to the product of the terms equidistant from it.

The correct answer is A (can be sides of a triangle) because the given condition implies that the square root of a, b, and c satisfy the triangle inequality.

Option D (12) is correct because it is the minimum value of 'a' when the given terms are in A.P., derived by using the AM-GM inequality or completing the square.

Option C (660) is correct because after simplifying the general term and summing up to the 10th term, the result is 660.

Option C ($\frac{{5\log 7 - 6\log 3}}{{\log 162 - \log 7}}$) is correct because taking the logarithm of both sides and solving for x results in the given expression.

Option A (9) is correct because it is the least value obtained by applying the AM-GM inequality.

Option C (2) is correct because the infinite sum of the reciprocals of the recursively defined sequence converges to 2.

Option C ($\frac{11}{2}$) is correct because the series can be expressed as a telescoping series, allowing for simplification and direct calculation.

Option D is correct because the sum of the series can be expressed and simplified to $\frac{5}{9}\left[n-\frac{1}{9}\left(1-\frac{1}{10^n}\right)\right]$.

Option C ($\frac{1}{12} n(n+1)(n^2-3 n+8)$) is correct because the sum of n terms of the series is derived to be in this form by finding the nth term and summing it.

Option C (30) is correct because it is the minimum value of the given expression, obtained by applying the AM-GM inequality to each factor.

Option A ($12 \mathrm{~cm}$) is correct because, given the volume and surface area, the longest edge of the rectangular solid block in GP is calculated to be 12 cm.

Option A (2007006) is correct because the sum of the series can be simplified by pairing terms and using the formula for the sum of integers.

Option B ($a^2+b-c=0$) is correct because when the given values are substituted and simplified, the equation holds true.

Option B (231) is correct because by factoring the given sum of squares and solving for the first term and common ratio, the sum of the first three terms of the GP is found to be 231.

Option B ($\frac{{n}}{{2}} \log \left(\frac{{a^{{n+3}}}}{{b^{{n-1}}}}\right)$) is correct. By using the properties of logarithms and summing the series, we get the result.

Option A ($n p(n p+1) / 2$) is correct because it is the sum of the sums of the first n terms of p arithmetic progressions, calculated using the given first terms and common differences.