Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
BITSAT2022MathematicsSequences and SeriesActual

Let a₁, a₂, a₄₀ be in AP and h₁, h₂, . h₁₀ be in HP. If a₁=h₁=2 and a₁₀=h₁₀=3 , then a₄ h₇ is

Options

  1. A2
  2. B3
  3. C5
  4. D6

Correct answer

D. 6

Step-by-step solution

Let d be the common difference of the AP. Then, aligned & a₁₀=3 a₁+9 d=3 & 2+9 d=3 d= 1 9 & a₄=a₁+3 d=2+ 1 3 = 7 3 aligned Let D be the common difference of 1 h₁ , 1 h₂ , .- 1 h₁₀ . Then, h₁₀=3 array ll & 1 h₁₀ = 1 3 1 2 +9 D= 1 3 & 9 D=- 1 6 D=- 1 54 & 1 h₇ = 1 h₁ +6 D= 1 2 - 1 9 = 7 18 & h₇= 18 7 & a₄ h₇= 7 3 18 7 =6 array

Practice Sequences and Series on Quantrex Academy →

More from Sequences and Series

If a 0, b 0, c 0 and a, b, c are distinct, then (a+b)(b+c)(c+a) is greater than 2024If _ k=1 ^n k(k+1)(k-1)=p n^4+q n^3+t n^2+s n where p, q, t and s are constants, then the value of s is equal to 2024There are four numbers of which the first three are in GP and the last three are in AP, whose common difference is 6 . If the first and the last numbers are equal, then two other n 2024If A=1+r^a+r^ 2 a +r^ 3 a + and B =1+ r ^ b + r ^ 2 ~b + r ^ 3 ~b + , then a b is equal to 2024The sum of the infinite series 1+ 5 6 + 12 6^2 + 22 6^3 + 35 6^4 + 51 6^5 + 70 6^6 + . . is equal to: 2024If ⁻¹ [ 1 1+1.2 ]+ ⁻¹ [ 1 1+2.3 ]+ + ⁻¹ [ 1 1+n(n+1) ]= ⁻¹[x] , then x is equal to 2024If arithmetic mean of two distinct positive real number a and b(a b) be twice their geometric mean, then a: b= 2024If y= ⁻¹ 1 x^2+x+1 + ⁻¹ 1 x^2+3 x+3 + ⁻¹ 1 x^2+5 x+7 + .to n terms, then d y d x = 2024 Full Sequences and Series list All BITSAT PYQs