Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Find the value of the sum of the 10th terms of the PRIME integers, where each term is defined as <img alt="\mathrm{4 n^2+5 n-1.}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B4%

Find the value of the sum of the 10th terms of the PRIME integers, where each term is defined as \mathrm{4 n^2+5 n-1.}

Option: 1

5068


Option: 2

6500


Option: 3

4035


Option: 4

9530


Answers (1)

best_answer

To find the sum of the 10th terms of the PRIME integers, where each term is defined as \mathrm{4 n^2+5 n-1}, we need to substitute the values of n from 1 to 10 into the expression and sum them up.

The sum of the first 10 terms can be calculated as follows:

\mathrm{ \text { Sum }=\left(4(1)^2+5(1)-1\right)+\left(4(2)^2+5(2)-1\right)+\ldots+\left(4(10)^2+5(10)-1\right) }

Simplifying the expression for each term, we have:

\mathrm{ \text { Sum }=(4+5-1)+(16+10-1)+\ldots+(400+50-1) }

Simplifying further, we get:

\mathrm{ \text { Sum }=8+25+\ldots+799 }

To find the sum of an arithmetic series, we can use the formula:

\mathrm{ \text { Sum }=(n / 2)(\text { first term }+ \text { last term }) }

In this case, the first term is 8 , the last term is 799 , and the number of terms is 10 . Plugging these values into the formula, we get:

\text { Sum }=(10 / 2)(8+799)=5(807)=4035

Therefore, the value of the sum of the 10th terms of the PRIME integers, where each term is defined as \mathrm{4 n^2+5 n-1}, is 4035 .

Posted by

manish painkra

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2026: JCECEB invites fresh applications for Jharkhand state merit list correction

Latest Question