Skip to main contentSkip to question navigationSkip to solution

IPMAT Indore 2021 (SA) PYQs

View all IPMAT/JIPMAT PYQs →
Collection of AfterBoards IPMAT JIPMAT Past Year Papers with Solutions

IPMAT Indore 2021

Algebra
>
Progression & Series

Medium

It is given that the sequence {xnx_n} satisfies x1=0,xn+1=xn+1+2√(1+xn)x_1 = 0, x_{n+1} = x_n + 1 + 2√(1+x_n) for n=1,2,...n = 1,2,... Then x31x_{31} is _______

Entered answer:

✅ Correct Answer: 960

Related questions:

IPMAT Indore 2019

There are numbers a1,a2,a3,…,ana_1, a_2, a_3, \ldots, a_n each of them being +1+1 or −1-1. If it is known that a1a2+a2a3+a3a4+…an−1an+ana1=0a_1 a_2 + a_2 a_3 + a_3 a_4 + \ldots a_{n-1} a_n + a_n a_1 = 0 then

IPMAT Indore 2019

Let α,β\alpha, \beta be the roots of x2−x+p=0x^2 - x + p = 0 and γ,δ\gamma, \delta be the roots of x2−4x+q=0x^2 - 4x + q = 0 where p and q are integers. If α,β,γ,δ\alpha, \beta, \gamma, \delta are in geometric progression then p+qp + q is

IPMAT Indore 2019

The number of terms common to both the arithmetic progressions 2,5,8,11,...,1792, 5, 8, 11, ..., 179 and 3,5,7,9,...,1013, 5, 7, 9, ..., 101 is

Keyboard Shortcuts

  • Left arrow: Previous question
  • Right arrow: Next question
  • S key: Jump to solution
  • Q key: Jump to question