BITSAT 2023 | Sequences and Series Question 7 | Mathematics

Given, a sequence of 4 numbers, first three of which are in GP and the last three are in AP with common difference 6. If first and last term of this sequence are equal, then the last term is

BITSAT 2022

MCQ (Single Correct Answer)

-1

Let a₁, a₂, ...... a₄₀ be in AP and h₁, h₂, ..... h₁₀ be in HP. If a₁ = h₁ = 2 and a₁₀ = h₁₀ = 3, then a₄h₇ is

BITSAT 2022

MCQ (Single Correct Answer)

-1

Let a₁, a₂, a₃ .... be a harmonic progression with a₁ = 5 and a₂₀ = 25. The least positive integer n for which a_n < 0, is

BITSAT 2022

MCQ (Single Correct Answer)

-1

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

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

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

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

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

Previous Next

Questions Asked from MCQ (Single Correct Answer)

BITSAT 2024 (4) BITSAT 2023 (3) BITSAT 2022 (3) BITSAT 2021 (2) BITSAT 2020 (5)

BITSAT Subjects

English Proficiency

Logical Reasoning

Verbal

Direction Sense Test Analogy Classification Series Completion Blood Relations Logical Deduction Logic Chart

Non Verbal

Completion of Figure Figure Formation Paper Folding and Cutting Figure Matrix Rule Detection Mirror Image