Sequence of series practice · Mixed difficulty

Option C ($2(d-c)$) is correct because in an arithmetic progression, the difference between consecutive terms is constant, so $e-c = 2(d-c)$.

Option D ($\frac{1}{{72}}$) is correct because the sum of the infinite series is found by expressing the general term as a difference of two terms and then summing the series.

Option B ($\frac{{n-(n+1) x+x^{{n+1}}}}{{(1-x)^2}}$) is correct because it is the result of summing the given series using the formula for a geometric series.

Option C ($\frac{\pi}{3}$) is correct because solving the GP condition leads to $\cos \theta = \frac{1}{2}$, which corresponds to $\theta = \frac{\pi}{3}$.

Option D (a composite number) is correct because the expression for Tr simplifies to a product of two integers, indicating a composite number.

Option B ($p q+r=p r+q r$) is correct because substituting the series representations of p, q, and r and simplifying leads to the equality.

Option B (7) is correct because solving the given equations for the sum of extreme terms and the product of middle terms yields the greatest number as 7.

Option D (None of these) is correct because solving the logarithmic equation yields $x = 0$, which is not among the provided options.

Option A (m+n) is correct because there are m constraints from the first condition and n constraints from the second condition, totaling m+n constraints.

Option B ($\left( {n - 1} \right){x_1}{x_n}$) is correct because the sum of the product of consecutive terms in a harmonic progression is equal to (n-1) times the product of the first and last terms.

Option B ($4^5$) is correct because the product of the first five terms of a G.P. with the third term being 4 is $4^5$.

Option B is correct because when x=0, the given equation simplifies to e^(e0) = e¹ = e, which equals a_0.

Option C ($3,-6,12,-24$) is correct because these numbers satisfy the given conditions for a GP.

Option G.P. is correct because if $b-c, 2b-x, b-a$ are in H.P., then $a-(x/2), b-(x/2), c-(x/2)$ are in G.P.

Option C ($\frac{20}{11}$) is correct because the series can be expressed as a telescoping series after simplifying the general term.

Option B 12 is correct because it is found by setting up an equation relating the sum of first n natural numbers to the sum of their cubes and solving for n.

Option C (HP) is correct because if a, b, c are in AP, then 1/(1-a), 1/(1-b), 1/(1-c) are in HP, which corresponds to x, y, and z.

Option B ($\frac{7}{29}$) is correct because the sum telescopes, leaving only the first and last terms, which simplify to the given fraction.

Option A (AP) is correct because the logarithm of terms in a GP, when raised to the same power, forms an AP.

Option C (64) is correct. The expression $\frac{{1 + n + {n^2} + ... + {n^{127}}}}{{{n^m} + 1}}$ is an integer when $m = 64$ because it allows for factorization using the difference of squares.

The correct answer is 3 because the series can be simplified by rationalizing each term, leading to a telescoping series where most terms cancel out, leaving only the first and last terms.

Option B (Assertion is correct, reason is correct; reason is not a correct explanation for assertion) is correct because while both statements are true, the reason (sum of infinite GP) doesn't explain the assertion (multiplying/dividing a GP).

Option A (2925) is the correct answer as it is the result of summing the squares of odd numbers from 1 to 25 using the formula for the sum of squares of an arithmetic progression.

Option C (1210) is correct because it is the sum of all two-digit numbers that leave a remainder of 1 when divided by 4.

Option D ($n - 1$) is correct because the series telescopes, leaving only the last term's square root minus the first term's square root.

Option A (5,20) is correct because the numbers 5 and 20 satisfy both the geometric mean (GM) and harmonic mean (HM) conditions.

Option D (1) is correct because by setting up equations based on the given information and solving, we find the least number in the AP is 1.

Option C (-4) is correct because by finding the common difference and using the properties of arithmetic progressions, the 20th term can be determined.

Option A (8/49) is correct because by finding the value of p and then calculating the 3rd and 4th terms, their sum is 8/49.

Option B is correct because the repeating decimal is converted into a fraction by subtracting the non-repeating part and dividing by the appropriate power of 10.