Q.If $\theta_1, \theta_2, \theta_3, \ldots, \theta_n$ are in A.P., whose common difference is $d$, show that $\sec\theta_1 \sec\theta_2 + \sec\theta_2 \sec\theta_3 + \ldots + \sec\theta_{n-1} \sec\theta_n = \dfrac{\tan\theta_n - \tan\theta_1}{\sin d}$.

Showing the 12 most recent of 18 on this concept.

1. AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
   Q.The value of $n$ for which the sum $\dfrac{{}^nC_0}{2^n} + 2.\dfrac{{}^nC_1}{2^n} + 3.\dfrac{{}^nC_2}{2^n} + \cdots + (n+1)\dfrac{{}^nC_n}{2^n} = 16$ is (A) $10$ (B) $20$ (C) $25$ (D) $30$

   ›Reveal solutionSolution

   The sum is the expected value of $(X+1)$ where $X \sim \text{Binomial}(n, 1/2)$, which equals $\frac{n}{2}+1$. Setting this equal to $16$ gives $n=30$, so the answer is option (D).

   We are asked to find $n$ such that

   $S = \frac{{}^nC_0}{2^n} + 2\cdot\frac{{}^nC_1}{2^n} + 3\cdot\frac{{}^nC_2}{2^n} + \cdots + (n+1)\frac{{}^nC_n}{2^n} = 16.$

   Concept and intuition: Sum of Products and Binomial Expectation

   Each term is of the form $(k+1) \cdot \frac{{}^nC_k}{2^n}$. Notice that $\frac{{}^nC_k}{2^n}$ is exactly the probability that a binomial random variable $X$ with $n$ trials and success probability $1/2$ equals $k$. So the sum is

   $S = \sum_{k=0}^n (k+1) \cdot P(X=k) = E[X+1] = E[X] + 1.$

   For a binomial distribution, $E[X] = n \cdot \frac12 = \frac{n}{2}$. Hence $S = \frac{n}{2} + 1$. Setting this equal to $16$ gives $\frac{n}{2} = 15$, so $n=30$.

   Now let’s verify step by step without relying on probability, using algebra.

   1. Write the sum in sigma notation

   $S = \sum_{k=0}^n (k+1) \frac{\binom{n}{k}}{2^n}.$

   The factor $2^n$ is constant, so

   $S = \frac{1}{2^n} \sum_{k=0}^n (k+1) \binom{n}{k}.$

   2. Split the sum

   $\sum_{k=0}^n (k+1) \binom{n}{k} = \sum_{k=0}^n k \binom{n}{k} + \sum_{k=0}^n \binom{n}{k}.$

   The second sum is the total number of subsets of an $n$-element set: $\sum_{k=0}^n \binom{n}{k} = 2^n$.

   3. Evaluate $\sum k \binom{n}{k}$ Use the identity $k \binom{n}{k} = n \binom{n-1}{k-1}$ (valid for $k \ge 1$). Then $\sum_{k=0}^n k \binom{n}{k} = \sum_{k=1}^n n \binom{n-1}{k-1} = n \sum_{j=0}^{n-1} \binom{n-1}{j} = n \cdot 2^{n-1}.$ …

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2. AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQ
   Q.Assertion (A): If $2 + 6 + 16 + 40 + \cdots$ to $k$ terms $= 4608$, then $k = 9$. Reason (R): $2 + 3\cdot 2 + 4\cdot 2^2 + \cdots$ to $n$ terms $= n\cdot 2^n\ \forall\, n \in N$. Which one of the following option is correct? (A) (A) and (R) are true and (R) is the correct explanation of (A) (B) (A) and (R) are true and (R) is not the correct explanation of (A) (C) (A) is true but (R) is false (D) (A) is false but (R) is true

   ›Reveal solutionSolution

   The series is $\sum (r+1)2^{r-1}$, whose sum to $n$ terms is exactly $n\cdot 2^n$ (R true); setting $k\cdot 2^k=4608$ gives $k=9$ (A true), and R is precisely the identity used, so it explains A.

   Identify the series. The terms $2,6,16,40,\dots$ are

   $2=2\cdot 2^0,\; 6=3\cdot 2^1,\; 16=4\cdot 2^2,\; 40=5\cdot 2^3,\dots$

   so the $r$-th term is $(r+1)2^{r-1}$, matching the Reason's series $2+3\cdot2+4\cdot2^2+\cdots$.

   Verify the Reason.

   $\sum_{r=1}^{n}(r+1)2^{r-1}=\sum_{r=1}^{n} r\,2^{r-1}+\sum_{r=1}^{n}2^{r-1}=\big[(n-1)2^{n}+1\big]+\big[2^{n}-1\big]=n\cdot 2^{n}.$

   So $2+3\cdot2+4\cdot2^2+\cdots$ to $n$ terms $=n\cdot 2^{n}$. R is true.

   Verify the Assertion. Using this result, the sum to $k$ terms equals $k\cdot 2^{k}$. Set it to $4608$: …

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3. AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQ
   Q.If $2.5+5.9+8.13+11.17+\ldots$ to n terms $=an^3+bn^2+cn+d$, then $a-b+c-d=$ (A) $7$ (B) $5$ (C) $-3$ (D) $-1$

   ›Reveal solutionSolution

   The general term is $(3r-1)(4r+1)=12r^2-r-1$; summing gives $S_n=4n^3+\tfrac{11}{2}n^2+\tfrac12 n$, so $a-b+c-d=-1$.

   Concept and Intuition

   The two interleaved factor sequences $2,5,8,11,\dots$ and $5,9,13,17,\dots$ are both arithmetic progressions, so the $r$-th term of the product series is a simple quadratic in $r$; summing a quadratic-in-$r$ series over $r=1$ to $n$ always gives a cubic polynomial in $n$ (via the standard $\sum r^2,\sum r,\sum 1$ formulas), matching the given form $an^3+bn^2+cn+d$.

   Step-by-Step Solution

   1. First factors: $2,5,8,11,\dots \Rightarrow$ $r$-th term $=3r-1$.
   2. Second factors: $5,9,13,17,\dots \Rightarrow$ $r$-th term $=4r+1$.
   3. General term: $T_r=(3r-1)(4r+1)=12r^2+3r-4r-1=12r^2-r-1$.
   4. Sum to $n$ terms: $S_n=12\sum r^2-\sum r-\sum 1 = 12\cdot\dfrac{n(n+1)(2n+1)}{6}-\dfrac{n(n+1)}{2}-n$.
   5. Simplify: $12\cdot\dfrac{n(n+1)(2n+1)}{6}=2n(n+1)(2n+1)=4n^3+6n^2+2n$.
   6. $S_n = 4n^3+6n^2+2n-\dfrac{n^2+n}{2}-n = \dfrac{8n^3+12n^2+4n-n^2-n-2n}{2}=\dfrac{8n^3+11n^2+n}{2}$.
   7. So $a=4,\ b=\dfrac{11}{2},\ c=\dfrac12,\ d=0$. …

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4. AP EAPCET 2025Set eng-2025-05-23-FN1 markMCQ
   Q.If $1, \omega, \omega^2$ are the cube roots of unity, then $1\left(2+\dfrac{1}{\omega}\right)\left(2+\dfrac{1}{\omega^2}\right) + 2\left(3+\dfrac{1}{\omega}\right)\left(3+\dfrac{1}{\omega^2}\right) + 3\left(4+\dfrac{1}{\omega}\right)\left(4+\dfrac{1}{\omega^2}\right) + \ldots\ 10\ \text{terms} =$ (A) $3080$ (B) $3465$ (C) $3175$ (D) $3715$

   ›Reveal solutionSolution

   Using $1/\omega=\omega^2$, $1/\omega^2=\omega$, $\omega+\omega^2=-1$, each bracketed pair simplifies to $n^2-n+1$; summing $r(r^2+r+1)$ for $r=1$ to $10$ gives $3465$. Answer: (B).

   Concept and Intuition

   The cube roots of unity satisfy $\omega^3=1$ and $1+\omega+\omega^2=0$, so $\omega+\omega^2=-1$ and $\omega\cdot\omega^2=\omega^3=1$. These two identities let any expression of the form $(n+\omega)(n+\omega^2)$ collapse to a clean polynomial in $n$: expanding gives $n^2+n(\omega+\omega^2)+\omega\cdot\omega^2 = n^2-n+1$. Recognising $1/\omega=\omega^2$ and $1/\omega^2=\omega$ (since $\omega\cdot\omega^2=1$) converts the given series into exactly this shape, after which the whole sum reduces to a routine sum of cubes/squares/naturals.

   Step-by-Step Solution

   1. Note $1/\omega=\omega^2$ and $1/\omega^2=\omega$ (because $\omega\cdot\omega^2=\omega^3=1$).
   2. The $r$-th term of the series is $r\left((r+1)+\dfrac1\omega\right)\left((r+1)+\dfrac1{\omega^2}\right) = r\big((r+1)+\omega^2\big)\big((r+1)+\omega\big)$.
   3. Let $n=r+1$. Expand $(n+\omega^2)(n+\omega) = n^2 + n(\omega+\omega^2) + \omega^3 = n^2 - n + 1$ (since $\omega+\omega^2=-1$, $\omega^3=1$).
   4. Substitute back $n=r+1$: $(r+1)^2-(r+1)+1 = r^2+2r+1-r-1+1 = r^2+r+1$. …

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5. AP EAPCET 2025Set eng-2025-05-27-FN1 markMCQ
   Q.If $\alpha,\beta$ are the roots of $x^2-5\gamma x-6\delta=0$ and $\gamma,\delta$ are the roots of $x^2-5\alpha x-6\beta=0$, then $\alpha+\beta+\gamma+\delta=$ (A) $0$ (B) $125$ (C) $144$ (D) $180$

   ›Reveal solutionSolution

   Using Vieta's formulas on both quadratics and eliminating variables shows $\alpha+\gamma=36$ (the genuine, non-degenerate case), so $\alpha+\beta+\gamma+\delta = 5(\alpha+\gamma)=180$.

   Concept and Intuition

   For each quadratic $x^2-(\text{sum})x+(\text{product})=0$, Vieta's formulas relate the roots' sum and product to the coefficients. Here we have two "coupled" quadratics — the coefficients of one involve the roots of the other — so writing down all four Vieta relations and eliminating variables lets us pin down the total sum $\alpha+\beta+\gamma+\delta$.

   Step-by-Step Solution

   1. $\alpha,\beta$ are roots of $x^2-5\gamma x-6\delta=0$, so by Vieta: $\alpha+\beta=5\gamma$ …(1), $\alpha\beta=-6\delta$ …(3).
   2. $\gamma,\delta$ are roots of $x^2-5\alpha x-6\beta=0$, so by Vieta: $\gamma+\delta=5\alpha$ …(2), $\gamma\delta=-6\beta$ …(4).
   3. The quantity sought, $S=\alpha+\beta+\gamma+\delta = (\alpha+\beta)+(\gamma+\delta) = 5\gamma+5\alpha = 5(\alpha+\gamma)$, so we just need $\alpha+\gamma$.
   4. From (1): $\beta = 5\gamma-\alpha$. From (2): $\delta=5\alpha-\gamma$.
   5. Substitute into (3): $\alpha(5\gamma-\alpha) = -6(5\alpha-\gamma) \Rightarrow 5\alpha\gamma-\alpha^2 = -30\alpha+6\gamma \Rightarrow 5\alpha\gamma-\alpha^2+30\alpha-6\gamma=0$. …(A)
   6. Substitute into (4): $\gamma(5\alpha-\gamma) = -6(5\gamma-\alpha) \Rightarrow 5\alpha\gamma-\gamma^2 = -30\gamma+6\alpha \Rightarrow 5\alpha\gamma-\gamma^2+30\gamma-6\alpha=0$. …(B)
   7. Subtract (B) from (A): $(-\alpha^2+30\alpha-6\gamma)-(-\gamma^2+30\gamma-6\alpha) = 0 \Rightarrow \gamma^2-\alpha^2+36\alpha-36\gamma=0$.
   8. Factor: $(\gamma-\alpha)(\gamma+\alpha) - 36(\gamma-\alpha) = 0 \Rightarrow (\gamma-\alpha)(\gamma+\alpha-36)=0$. …

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6. TG EAPCET 2024Set eng-2024-05-09-AN1 markMCQ
   Q.If $\alpha, \beta, \gamma$ are the roots of the equation $2x^3 - 5x^2 + 4x - 3 = 0$, then $\sum \alpha \beta (\alpha + \beta) =$ (A) $8$ (B) $4$ (C) $2$ (D) $\dfrac{1}{2}$

   ›Reveal solutionSolution

   Using $\sum\alpha\beta(\alpha+\beta) = (\sum\alpha)(\sum\alpha\beta) - 3\alpha\beta\gamma = \tfrac52\cdot 2 - 3\cdot\tfrac32 = \tfrac12$.

   For $2x^3 - 5x^2 + 4x - 3 = 0$, Vieta's formulas give

   $\alpha+\beta+\gamma = \frac{5}{2}, \qquad \alpha\beta+\beta\gamma+\gamma\alpha = \frac{4}{2} = 2, \qquad \alpha\beta\gamma = \frac{3}{2}.$

   The required symmetric sum expands as …

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7. AP EAPCET 2024Set eng-2024-05-19-AN1 markMCQ
   Q.$2.5 + 5.9 + 8.13 + 11.17 + \cdots$ to 10 terms $=$ (A) 3355 (B) 4555 (C) 1375 (D) 1380

   ›Reveal solutionSolution

   Each term is a product of the $n$-th terms of two APs; expanding and summing standard formulas for $\sum n^2$ and $\sum n$ over 10 terms gives $4555$ — option (B).

   Concept and Intuition

   The given series $2\cdot5 + 5\cdot9 + 8\cdot13 + 11\cdot17+\cdots$ pairs terms from two separate arithmetic progressions:

   • First factor: $2,5,8,11,\ldots$ — AP with first term $2$, common difference $3$, so $n$-th term $=2+3(n-1)=3n-1$.
   • Second factor: $5,9,13,17,\ldots$ — AP with first term $5$, common difference $4$, so $n$-th term $=5+4(n-1)=4n+1$.

   The general term of the series is the product $T_n=(3n-1)(4n+1)$, a quadratic in $n$, so the sum can be evaluated using the standard formulas $\sum_{n=1}^{N}n = \dfrac{N(N+1)}2$ and $\sum_{n=1}^N n^2=\dfrac{N(N+1)(2N+1)}6$.

   Step-by-Step Solution

   1. Confirm the pattern: $T_1=2\cdot5=10$, $T_2=5\cdot9=45$, $T_3=8\cdot13=104$, $T_4=11\cdot17=187$ — matches $(3n-1)(4n+1)$ for $n=1,2,3,4$.
   2. Expand: $T_n=(3n-1)(4n+1)=12n^2+3n-4n-1=12n^2-n-1$. …

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8. AP EAPCET 2024Set eng-2024-05-20-FN1 markMCQ
   Q.$\frac{1}{3\cdot 7} + \frac{1}{7\cdot 11} + \frac{1}{11\cdot 15} + \cdots$ to $50$ terms $=$ (A) $\dfrac{50}{203}$ (B) $\dfrac{50}{609}$ (C) $\dfrac{150}{203}$ (D) $\dfrac{25}{609}$

   ›Reveal solutionSolution

   This is a telescoping series in disguise: partial-fraction each term and almost everything cancels, leaving $\dfrac{50}{609}$.

   Concept and Intuition

   Whenever consecutive terms' denominators share a common difference (here, factors $3,7,11,15,\ldots$ each differing by $4$), splitting each term via partial fractions turns the sum into a telescoping one, where interior terms cancel and only the first and last survive.

   Step-by-Step Solution

   1. The $k$-th term (for $k=1,2,\ldots$) is $\dfrac{1}{(4k-1)(4k+3)}$: check $k=1$ gives $\dfrac{1}{3\cdot7}$, $k=2$ gives $\dfrac{1}{7\cdot11}$, etc. — matches the pattern.
   2. Partial fractions: $\dfrac{1}{(4k-1)(4k+3)} = \dfrac14\left(\dfrac{1}{4k-1}-\dfrac{1}{4k+3}\right)$, since the difference between the two denominators is $4$.
   3. Summing from $k=1$ to $50$:

   $\sum_{k=1}^{50}\frac14\left(\frac{1}{4k-1}-\frac{1}{4k+3}\right) = \frac14\left[\left(\frac13-\frac17\right)+\left(\frac17-\frac{1}{11}\right)+\cdots+\left(\frac{1}{199}-\frac{1}{203}\right)\right]$

   Every interior term cancels (telescopes), leaving only the very first and very last pieces: …

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9. TG EAPCET 2023Set eng-2023-05-12-AN1 markMCQ
   Q.If $\alpha$ and $\beta$ are the roots of the equation $x^2 + 2x + 2 = 0$, then $\alpha^{15} + \beta^{15} =$ (A) $-512$ (B) $-256$ (C) $256$ (D) $512$

   ›Reveal solutionSolution

   The roots are complex conjugates with modulus $\sqrt{2}$ and argument $\frac{3\pi}{4}$. Using De Moivre’s theorem, $\alpha^{15}+\beta^{15}=2(\sqrt{2})^{15}\cos\left(15\cdot\frac{3\pi}{4}\right) = -256$.

   The equation $x^2 + 2x + 2 = 0$ has discriminant $4 - 8 = -4$, so its roots are complex. The key insight is that when a quadratic has complex conjugate roots, powers of those roots can be handled elegantly using polar form and De Moivre’s theorem — no need to compute each power separately.

   Let’s find the roots first. Using the quadratic formula:

   $x = \frac{-2 \pm \sqrt{4 - 8}}{2} = \frac{-2 \pm 2i}{2} = -1 \pm i$

   So $\alpha = -1 + i$ and $\beta = -1 - i$ (or vice versa; the sum is symmetric).

   Now convert $-1 + i$ to polar form. Its modulus is $\sqrt{(-1)^2 + 1^2} = \sqrt{2}$. Its argument: the point $(-1, 1)$ lies in the second quadrant, so $\theta = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$. Thus:

   $\alpha = \sqrt{2} \, e^{i\frac{3\pi}{4}}, \quad \beta = \sqrt{2} \, e^{-i\frac{3\pi}{4}}$

   1. Apply De Moivre’s theorem to each:

   $\alpha^{15} = (\sqrt{2})^{15} \, e^{i\cdot 15\cdot\frac{3\pi}{4}} = 2^{15/2} \, e^{i\frac{45\pi}{4}}$

   $\beta^{15} = 2^{15/2} \, e^{-i\frac{45\pi}{4}}$

   2. Simplify the angle: $\frac{45\pi}{4} = 11\pi + \frac{\pi}{4}$. Since $e^{i(11\pi + \pi/4)} = e^{i11\pi} e^{i\pi/4} = e^{i\pi} e^{i\pi/4}$ (because $11\pi$ is an odd multiple of $\pi$, so $e^{i11\pi} = -1$). Thus:

   $e^{i\frac{45\pi}{4}} = - e^{i\pi/4}$

   Similarly, $e^{-i\frac{45\pi}{4}} = - e^{-i\pi/4}$.

   3. Add them:

   $\alpha^{15} + \beta^{15} = 2^{15/2} \left( - e^{i\pi/4} - e^{-i\pi/4} \right) = -2^{15/2} \cdot 2\cos\frac{\pi}{4}$

   Since $\cos\frac{\pi}{4} = \frac{\sqrt{2}}{2}$, we get:

   $$$$

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10. TG EAPCET 2023Set eng-2023-05-13-FN1 markMCQ
    Q.If the sum of two particular roots of the equation $x^4 - 4x^3 - 7x^2 + 22x + 24 = 0$ is equal to the sum of the remaining two roots, then the sum of the cubes of all the roots of this equation is (A) $100$ (B) $196$ (C) $0$ (D) $82$

    ›Reveal solutionSolution

    Using Vieta’s formulas and the given condition that the sum of two roots equals the sum of the other two, we deduce the roots are symmetric in pairs, which lets us compute the sum of cubes directly as $0$.

    We are given the quartic

    $x^4 - 4x^3 - 7x^2 + 22x + 24 = 0$

    and told that the sum of two particular roots equals the sum of the remaining two roots. Let the roots be $p, q, r, s$. The condition says

    $p+q = r+s.$

    ---

    1. Use Vieta’s formulas for the sum of all roots

    From the polynomial, the coefficient of $x^3$ is $-4$, so

    $p+q+r+s = 4.$

    Since $p+q = r+s$, let $S = p+q = r+s$. Then

    $S + S = 4 \quad\Rightarrow\quad 2S = 4 \quad\Rightarrow\quad S = 2.$

    So each pair sums to $2$. That is,

    $p+q = 2,\qquad r+s = 2.$

    ---

    2. Express the sum of cubes in terms of symmetric sums

    We want

    $p^3+q^3+r^3+s^3.$

    A useful identity:

    $p^3+q^3 = (p+q)^3 - 3pq(p+q).$

    Similarly for $r,s$. So

    $p^3+q^3+r^3+s^3 = (p+q)^3 + (r+s)^3 - 3pq(p+q) - 3rs(r+s).$

    Since $p+q = r+s = 2$, this becomes

    $= 2^3 + 2^3 - 3pq(2) - 3rs(2) = 8+8 - 6pq - 6rs = 16 - 6(pq+rs).$

    ---

    3. Find $pq+rs$ using Vieta’s other relations

    Vieta also gives:

    • Sum of products two at a time:

    $pq + pr + ps + qr + qs + rs = -7.$

    Group terms cleverly:

    $(pq+rs) + (pr+ps+qr+qs) = -7.$

    Factor the second group:

    $$$$

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11. AP EAPCET 2023Set eng-2023-05-16-FN1 markMCQ
    Q.If $3\times {}^5C_0 + 8\times {}^5C_1 + 13\times {}^5C_2 + 18\times {}^5C_3 + 23\times {}^5C_4 + 28\times {}^5C_5 = k\times 2^4$, then $k =$ (A) $33$ (B) $37$ (C) $31$ (D) $30$

    ›Reveal solutionSolution

    Splitting the AP coefficients $3+5i$ into two standard binomial sums ($\sum\binom5i=2^5$ and $\sum i\binom5i=5\cdot2^4$) gives $496=16k$, so $k=31$.

    Concept and Intuition

    When binomial coefficients $\binom ni$ are weighted by a linear function of $i$ (here $3+5i$), split the sum into the two standard identities: $\sum_i\binom ni=2^n$ and $\sum_i i\binom ni = n\cdot2^{n-1}$ (the second follows from $i\binom ni = n\binom{n-1}{i-1}$).

    Step-by-Step Solution

    1. Note the coefficients $3,8,13,18,23,28$ increase by $5$ each time, so the $i$-th term's coefficient (for $\binom5i$, $i=0,\ldots,5$) is $3+5i$.
    2. Sum $=\sum_{i=0}^{5}(3+5i)\binom5i = 3\sum_{i=0}^5\binom5i + 5\sum_{i=0}^5 i\binom5i$.
    3. $\sum_{i=0}^5\binom5i = 2^5=32$. …

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12. AP EAPCET 2023Set eng-2023-05-18-AN1 markMCQ
    Q.The sum of the fourth powers of the roots of the equation $16x^2-10x+1=0$ is (A) $\dfrac{257}{4096}$ (B) $\dfrac{257}{2048}$ (C) $\dfrac{257}{1024}$ (D) $\dfrac{257}{512}$

    ›Reveal solutionSolution

    Build up the fourth-power symmetric sum from the elementary symmetric sum and product of the roots using Newton-identity-style squaring.

    Concept and Intuition

    For a quadratic $ax^2+bx+c=0$ with roots $r_1,r_2$: $r_1+r_2=-b/a$, $r_1r_2=c/a$. Power sums of roots can be built up algebraically: $r_1^2+r_2^2=(r_1+r_2)^2-2r_1r_2$, and then $r_1^4+r_2^4=(r_1^2+r_2^2)^2-2(r_1r_2)^2$ — squaring twice rather than solving for the (irrational) roots directly.

    Step-by-Step Solution

    1. For $16x^2-10x+1=0$: sum $S=r_1+r_2=10/16=5/8$; product $P=r_1r_2=1/16$.
    2. $r_1^2+r_2^2 = S^2-2P = \dfrac{25}{64}-\dfrac{2}{16} = \dfrac{25}{64}-\dfrac{8}{64}=\dfrac{17}{64}$.
    3. $r_1^4+r_2^4 = (r_1^2+r_2^2)^2 - 2(r_1r_2)^2 = \left(\dfrac{17}{64}\right)^2 - 2\left(\dfrac{1}{16}\right)^2$. …

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