TS EAMCET PYQs for Binomial theorem with Solutions: Practice TS EAMCET Previous Year Questions

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Yashodeep Mahurkar

Updated on - Dec 20, 2025

Binomial theorem is an important topic in the Mathematics section in TS EAMCET exam. Practising this topic will increase your score overall and make your conceptual grip on TS EAMCET exam stronger.

This article gives you a full set of TS EAMCET PYQs for Binomial theorem with explanations for effective preparation. Practice of TS EAMCET Mathematics PYQs including Binomial theorem questions regularly will improve accuracy, speed, and confidence in the TS EAMCET 2026 exam.

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TS EAMCET PYQs for Binomial theorem with Solutions

  • 1.

    The numerically greatest term in the expansion of (3x - 16y)15 when x = 23 and y = 32 is ?

      • 13
         

      • 14
         

      • 15
         

      • 16


    • 2.

      If n is a positive integer and f(n) is the coeffcient of xn in the expansion of (1 + x)(1-x)n, then f(2023) =

        • -2021

        • 2022

        • 2023

        • -2023


      • 3.
        Numerically greatest term in the expansion of $(2x-3y)^n$ when $x=\frac{7}{5}, y=\frac{3}{7}$ and $n=13$ is

          • $13.3^5.7^9$
          • $13.3^4.7^9$
          • $26.3^5.7^9$
          • $26.3^4.7^9$

        • 4.
          If the coefficients of three consecutive terms in the expansion of (1 + x)23 are in arithmetic progression, then those terms are ?

            • \(T_{10},\,T_{11},\,T_{12}\)
            • \(T_{8},\,T_{9},\,T_{10}\)
            • \(T_{13},\,T_{14},\,T_{15}\)
            • \(T_{14},\,T_{15},\,T_{16}\)

          • 5.
            If $C_0, C_1, C_2, \dots, C_n$ are the binomial coefficients in the expansion of $(1+x)^n$ then $\sum_{r=1}^{n} \frac{r C_r}{C_{r-1}} =$

              • 540
              • 336
              • 105
              • 270

            • 6.

              If y = \(\frac{3}{4} + \frac{3.5}{4.8}+\frac{5.5.7}{4.8.12}+ \).... to ∞, then

                • y2 - 2y + 5 = 0

                • y2 + 2y - 7 = 0

                • y2 - 3y + 4 = 0

                • y2 + 4y - 6 = 0

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