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IPMAT Indore 2023 (SA) PYQs

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IPMAT Indore 2023

Algebra
>
Progression & Series

Hard

If f(n)=1+2+3+⋯+(n+1)f(n)= 1 + 2 + 3 +\cdots+(n+1) and g(n)=∑k=1k=n1f(k)g(n)= \sum_{k=1}^{k=n} \dfrac{1}{f(k)}, then the least value of nn for which g(n)g(n) exceeds the value 99100\dfrac{99}{100} is:

Entered answer:

✅ Correct Answer: 199

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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

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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

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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

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