BITSAT 2022 | Sequences and Series Question 10 | Mathematics

In a sequence of 21 terms, the first 11 terms are in AP with common difference 2 and the last 11 terms are in GP with common ratio 2. If the middle term of AP be equal to the middle term of the GP, then the middle term of the entire sequence is

$$-\frac{10}{31}$$

$$\frac{10}{31}$$

$$\frac{32}{31}$$

$$-\frac{31}{32}$$

BITSAT 2021

MCQ (Single Correct Answer)

-1

If a + 2b + 3c = 12, (a, b, c $$\in$$R⁺), then the maximum value of ab²c³ is

2³

2⁴

2⁶

2⁵

BITSAT 2021

MCQ (Single Correct Answer)

-1

Sum of n terms of the infinite series

1.3² + 2.5² + 3.7² + ..... $$\infty$$ is

$${n \over 6}(n + 1)(6{n^2} + 14n + 7)$$

$${n \over 6}(n + 1)(6{n^2} + 14n + 5)$$

$${n \over 6}(n + 1)(2n + 1)(3n + 1)$$

$$4{n^3} + 4{n^2} + n$$

BITSAT 2020

MCQ (Single Correct Answer)

-1

If a₁, a₂, a₃, ......., a₂₀ are AM's between 13 and 67, then the maximum value of a₁, a₂, a₃, ......, a₂₀ is equal to

(20)²⁰

(40)²⁰

(60)²⁰

(80)²⁰

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