Correct Answer: (13, 5, 21)

Solution : Given:
(6, 4, 8); (17, 6, 28)

Add the difference of the first and second numbers to the first number to get the third number –
⇒(6, 4, 8)→6 + (6 – 4) = 6 + 2 = 8
⇒(17, 6, 28)→17 +

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The final session results and the JEE Main 2024 cutoff will be made available on jeemain.nta.nic.in by NTA. To estimate their estimated qualifying marks for JEE Main 2024, candidates can consult the JEE Main cutoff from the previous year. The minimum score needed in JEE Mains to be

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School admissions generally start around the beginning of the calendar year that is in the months of January and February. However, the exact dates may vary as the school may change it according to their preference. So It's advisable that you should connect with the school through call or

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The eligibility criteria for BSc Nautical Science is that the candidates must have passed 10+2 from the Science stream with Physics, Chemistry, Mathematics, and English as mandatory subjects. Candidates must have completed their 10+2 with at least 60% aggregate marks.

All candidates applying for B.Sc Nautical Science Course also

Correct Answer: Kyzyl-Kum Desert

Solution : The correct option is - Kyzyl-Kum Desert.

The largest gold deposit is in KyzylKum. Uzbekistan's Kyzylkum Desert is located southeast of the Aral Sea and spans an area of around 115,000 square miles (300,000 square kilometers). It is situated between the Syr and Amu

Correct Answer: $\frac{k^{32}-\frac{1}{k^{32}}}{k+\frac{1}{k}}$

Solution : Consider, $\left(k-\frac{1}{k}\right)\left(k^2+\frac{1}{k^2}\right)\left(k^4+\frac{1}{k^4}\right)\left(k^8+\frac{1}{k^8}\right)\left(k^{16}+\frac{1}{k^{16}}\right)$
Multiplying and dividing by $(k+\frac{1}{k})$
⇒ $\left(k-\frac{1}{k}\right)\left(k^2+\frac{1}{k^2}\right)\left(k^4+\frac{1}{k^4}\right)\left(k^8+\frac{1}{k^8}\right)\left(k^{16}+\frac{1}{k^{16}}\right)=\frac{(k-\frac{1}{k})(k+\frac{1}{k})(k^2+\frac{1}{k^2})(k^4+\frac{1}{k^4})(k^8+\frac{1}{k^8})(k^{16}+\frac{1}{k^{16}})}{k+\frac{1}{k}}$
Now using, $(a+b)(a-b)=a^2 - b^2$
⇒ $\left(k-\frac{1}{k}\right)\left(k^2+\frac{1}{k^2}\right)\left(k^4+\frac{1}{k^4}\right)\left(k^8+\frac{1}{k^8}\right)\left(k^{16}+\frac{1}{k^{16}}\right)=\frac{(k^2-\frac{1}{k^2})(k^2+\frac{1}{k^2})(k^4+\frac{1}{k^4})(k^8+\frac{1}{k^8})(k^{16}+\frac{1}{k^{16}})}{k+\frac{1}{k}}$
$=\frac{(k^4-\frac{1}{k^4})(k^4+\frac{1}{k^4})(k^8+\frac{1}{k^8})(k^{16}+\frac{1}{k^{16}})}{k+\frac{1}{k}}$
$=\frac{(k^8-\frac{1}{k^8})(k^8+\frac{1}{k^8})(k^{16} + \frac{1}{k^{16}})}{k+\frac{1}{k}}$
$=\frac{(k^{16} - \frac{1}{k^{16}})(k^{16} + \frac{1}{k^{16}})}{k+\frac{1}{k}}$
$=\frac{k^{32} - \frac{1}{k^{32}}}{k+\frac{1}{k}}$
Hence, the correct answer is $\frac{k^{32} - \frac{1}{k^{32}}}{k+\frac{1}{k}}$.

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Yes, definitely you can get admission in Indian Institute of Mass Communication after graduating from YCMOU open university.

You have to clear CUET exam for being eligible in Institute of Mass Communication as these institute offers seat to students in the basis of score of CUET and personal