Binomial Theorem PYQs - Last 5 Years
AP EAPCET / Mathematics / Algebra / 46 recent questions
MathematicsAlgebra2021-2025
Practice 46 AP EAPCET Mathematics questions from Binomial Theorem. Use the year-wise and type-wise breakdown to prioritize recent PYQs, then continue into the question list below.
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Last 5 Years Binomial Theorem Questions
Showing 46 of 46 filtered questions.
1Binomial Theorem
If $x$ is positive real number and the first negative term in the expansion of $(1+x)^{\frac{27}{5}}$ is $t_k$, then $k=$
MCQ+1 / -02025
2Binomial Theorem
\(\frac{1}{81^n}-{ }^{2 n} C_1 \frac{10}{81^n}+{ }^{2 n} C_2 \frac{10^2}{81^n}-\ldots+\frac{10^{2 n}}{81^n}=\)
MCQ+1 / -02025
3Binomial Theorem
\(\sum_{r=1}^{15} r^2\left(\frac{{ }^{15} C_r}{{ }^{15} C_{r-1}}\right)=\)
MCQ+1 / -02025
4Binomial Theorem
The remainder obtained when $(2 m+1)^{2 n}(m, n \in N)$ is divided by 8 is
MCQ+1 / -02025
5Binomial Theorem
The numerically greatest term in the expansion of $(x+3 y)^{13}$, when $x=\frac{1}{2}$ and $y=\frac{1}{3}$ is
MCQ+1 / -02025
6Binomial Theorem
In the binomial expansion of $(p-q)^{14}$, if the sum of 7th term and 8 th term is zero, then $\frac{p+q}{p-q}=$
MCQ+1 / -02025
7Binomial Theorem
The coefficient of $x^{10}$ in the expansion of $\left(x+\frac{2}{x}-5\right)^{12}$ is
MCQ+1 / -02025
8Binomial Theorem
Let $S_1=\sum\limits_{j=1}^{10} j(j-1) \cdot{ }^{10} C_j, S_2=\sum\limits_{j=1}^{10} j \cdot{ }^{10} C_j$ and
\(S_3=\sum\limits_{j=1}^{10} j^2 \cdot{ }^{10} C_j\)
Assertion (A) $S_3=55 \times 2^9$
Reason (R) $S_1=90 \times 2^8$ and $S_2=1...
\(S_3=\sum\limits_{j=1}^{10} j^2 \cdot{ }^{10} C_j\)
Assertion (A) $S_3=55 \times 2^9$
Reason (R) $S_1=90 \times 2^8$ and $S_2=1...
MCQ+1 / -02025
9Binomial Theorem
Sum of the coefficients of $x^4$ and $x^6$ in the expansion of $\left(1+x-x^2\right)^6$ is
MCQ+1 / -02025
10Binomial Theorem
If $y=\frac{3}{4}+\frac{3 \cdot 5}{4 \cdot 8}+\frac{3 \cdot 5 \cdot 7}{4 \cdot 8 \cdot 12}+\ldots+\infty$, then
MCQ+1 / -02025
11Binomial Theorem
If the coefficients of $x^{10}$ and $x^{11}$ in the expansion of $\left(1+\alpha x+\beta x^2\right)(1+x)^{11}$ are 396 and 144 respectively, then $\alpha^2+\beta^2=$
MCQ+1 / -02025
12Binomial Theorem
If $-\frac{2}{3} < x < \frac{2}{3}$, then the value of the 5 th term in the expansion of $\frac{1}{\sqrt[3]{2-3 x}}$ when $x=\frac{1}{2}$ is
MCQ+1 / -02025
13Binomial Theorem
The mean and variance of a binomial distribution are $x$ and 5 respectively. If $x$ is an integer, then the possible values for $x$ are
MCQ+1 / -02025
14Binomial Theorem
If $C_0, C_2, \ldots, C_n$ are the binomial coefficients in the expansion of $(1+x)^n$, then
\(\left(C_0+C_1\right)-\left(C_2+C_3\right)+\left(C_4+C_5\right)-\left(C_6+C_7\right)+\ldots=\)
\(\left(C_0+C_1\right)-\left(C_2+C_3\right)+\left(C_4+C_5\right)-\left(C_6+C_7\right)+\ldots=\)
MCQ+1 / -02025
15Binomial Theorem
If $11^{12}-11^2=k\left(5 \times 10^9+6 \times 10^9+33 \times 10^8\right. \left.+110 \times 10^7+\ldots+33\right)$, then $k=$
MCQ+1 / -02025
16Binomial Theorem
If $k$ is a positive integer and $10^k$ is a divisor of the number $9^{11}+11^9$, then the greatest value of $k$ is
MCQ+1 / -02025
17Binomial Theorem
The number of all possible values of $k$ for which the expansion $(\sqrt{x}+\sqrt[k]{y})^{10}$ will have exactly nine irrational terms is
MCQ+1 / -02025
18Binomial Theorem
The coefficient of $x^3$ in the power series expansion of $\frac{1+4 x-3 x^2}{(1+3 x)^3}$ is
MCQ+1 / -02025
19Binomial Theorem
The terms containing $x^r y^s$ (for certain $r$ and $s$ ) are present in both the expansions of $\left(x+y^2\right)^{13}$ and $\left(x^2+y\right)^{14}$. If $\alpha$ is the number of such terms, then the $\operatorname{sum} \alpha \sum_{r, s...
MCQ+1 / -02025
20Binomial Theorem
The coefficient of $x^3$ in the expansion of $\frac{x^4+1}{\left(x^2+1\right)(x-1)}$ when it is expressed in terms of positive integral powers of $x$, is
MCQ+1 / -02025
21Binomial Theorem
Coefficient of $x^2$ in the expansion of $\left(x^2+x-2\right)^5$ is
MCQ+1 / -02025
22Binomial Theorem
If $P_n$ denotes the product of the binomial coefficients in the expansion of $(1+x)^n$, then $\frac{P_{n+1}}{P_n}=$
MCQ+1 / -02025
23Binomial Theorem
If $x$ is so large that terms containing $x^{-3}, x^{-4}, x^{-5}, \ldots$ can be neglected, then the approximate value of $\left(\frac{3 x-5}{4 x^2+3}\right)^{-1 / 5}$ is
MCQ+1 / -02025
24Binomial Theorem
If $(1+x)^n=\sum_{r=0}^n C, x^r$, then the value of $C_0+\left(C_0+C_1\right)+\left(C_0+C_1+C_2\right)+\ldots+ \left(C_0+C_1+C_2+\ldots+C_n\right)$ is
MCQ+1 / -02025
25Binomial Theorem
For $|x|<\frac{1}{\sqrt{2}}$, the coefficient of $x$ in the expansion of $\frac{(1-4 x)^2\left(1-2 x^2\right)^{1 / 2}}{(4-x)^{3 / 2}}$ is
MCQ+1 / -02024
26Binomial Theorem
The independent term in the expansion of $\left(1+x+2 x^2\right)\left(\frac{3 x^2}{2}-\frac{1}{3 x}\right)^9$ is
MCQ+1 / -02024
27Binomial Theorem
If $|x|<1$, then the number of terms in the expansion of $\left[\frac{1}{2}\left(1 \cdot 2+2 \cdot 3 x+3 \cdot 4 x^2+\ldots . \infty\right)\right]^{-25}$
MCQ+1 / -02024
28Binomial Theorem
If the $2 \mathrm{nd}, 3 \mathrm{rd}$ and 4 th terms in the expansion of $(x+a)^n$ are $96,216,216$ respectively and $n$ is a positive integer, then $a+x=$
MCQ+1 / -02024
29Binomial Theorem
If the coefficients of $(2 r+6)$ th and $(r-1)$ th terms in the expansion of $(1+x)^{21}$ are equal, then the value of $r$ is equal to
MCQ+1 / -02024
30Binomial Theorem
If $P$ is the greatest divisor of $49^n+16 n-1$ for all $n \in N$, then the number of factors of $P$ is
MCQ+1 / -02024
31Binomial Theorem
If the coefficients of $r$ th, $(r+1)$ th and $(r+2)$ th terms in the expansion of $(1+x)^n$ are in the ratio of $4: 15: 42$, then $n-r$ is equal to
MCQ+1 / -02024
32Binomial Theorem
The sum of the rational terms in the binomial expansion of $\left(\sqrt{2}+3^{1 / 5}\right)^{10}$ is
MCQ+1 / -02024
33Binomial Theorem
If the eleventh term in the binomial expansion of $(x+a)^{15}$ is the geometric mean of the eighth and twelfth terms, then the greatest term in the expansion is
MCQ+1 / -02024
34Binomial Theorem
If the ratio of the terms equidistant from the middle term in the expansion of $(l+x)^{12}$ is $\frac{1}{256}(x \in N)$, then sum of all the terms of the expansion $(1+x)^{12}$ is
MCQ+1 / -02024
35Binomial Theorem
The coefficient of $x^5$ in the expansion of $\left(2 x^3-\frac{1}{3 x^2}\right)^5$ is
MCQ+1 / -02024
36Binomial Theorem
If $|x|<\frac{2}{3}$, then the 4th term in the expansion of $(3 x-2)^{\frac{2}{3}}$ is :
MCQ+1 / -02024
37Binomial Theorem
If the coefficients of $x^5$ and $x^6$ are equal in the expansion of $\left(a+\frac{x}{5}\right)^{65}$, then the coefficient of $x^2$ in the expansion of $\left(a+\frac{x}{5}\right)^4$ is.
MCQ+1 / -02024
38Binomial Theorem
Numerically greatest term in the expansion of $(5+3 x)^6$ When, $x=1$, is
MCQ+1 / -02024
39Binomial Theorem
The absolute value of the difference of the coefficients of $x^4$ and $x^6$ in the expansion of $x^2 - 2x^2 + (x + 1)^4(x^2 - 1)^2$, is
MCQ+1 / -02024
40Binomial Theorem
The coefficient of $x^5$ in $\left(3+x+x^2\right)^6$ is
MCQ+1 / -02024
41Binomial Theorem
The square root of independent term in the expansion of $ \left( 2x^2 + \frac{5}{x} \right)^5 $ is
MCQ+1 / -02024
42Binomial Theorem
If $C_j$ stands for ${ }^n C_j$, then
\(\frac{C_1}{C_0}+\frac{2 \times C_2}{C_1}+\frac{3 \times C_3}{C_2}+\ldots+\frac{n \times C_n}{C_{n-1}}=\)
\(\frac{C_1}{C_0}+\frac{2 \times C_2}{C_1}+\frac{3 \times C_3}{C_2}+\ldots+\frac{n \times C_n}{C_{n-1}}=\)
MCQ+1 / -02023
43Binomial Theorem
If $(2-5 x)^{\frac{-1}{5}}=a_0+a_1 x+a_2 x^2+\ldots$, then $\frac{a_1}{a_2}=$
MCQ+1 / -02023
44Binomial Theorem
If $C_j={ }^n C_j$, then $C_0 C_r+C_1 C_{r+1}+C_2 C_{r+2}+\ldots+C_{n-r} C_n=$
MCQ+1 / -02023
45Binomial Theorem
The coefficient of the highest power of $x$ in the expansion of $\left(x+\sqrt{x^2-1}\right)^8+\left(x-\sqrt{x^2-1}\right)^8$ is
MCQ+1 / -02023
46Binomial Theorem
The least value of \(n\) so that \({ }^{(n-1)} C_3+{ }^{(n-1)} C_4>{ }^n C_3\)
MCQ+1 / -02022
