Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Compute the sum of the 10 th terms of the PRIME integers, where the nth term is given by <img alt="\mathrm{n^2+3 n-2.}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7Bn%5E2&pl

Compute the sum of the 10 th terms of the PRIME integers, where the nth term is given by \mathrm{n^2+3 n-2.}

Option: 1

900


Option: 2

650


Option: 3

370


Option: 4

410


Answers (1)

best_answer

To find the sum of the 10 th terms of the sequence generated by the expression n^2+3 n-2. 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:

\text { Sum }=\left(1^2+3(1)-2\right)+\left(2^2+3(2)-2\right)+\ldots+\left(10^2+3(10)-2\right)
Simplifying the expression for each term, we have:

\text { Sum }=(1+3-2)+(4+6-2)+\ldots+(100+30-2)
Simplifying further, we get:

\mathrm{ \text { Sum }=2+8+\ldots+128 }
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 2, the last term is 128 , and the number of terms is 10 .

Plugging these values into the formula, we get:

\mathrm{ \operatorname{Sum}=(10 / 2)(2+128)=5(130)=650 }

Therefore, the sum of the 10 th terms of the sequence generated by  \mathrm{n^2+3 n-2 \text {. is } 650 \text {. }}

Posted by

Gunjita

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: NTA reopens Session 2 registration; apply by March 13 at jeemain.nta.nic.in
anam-khan

JEE Mains 2026 LIVE: NTA session 2 city slip soon; admit card, exam dates, here's how to prepare
anam-khan

JEE Main 2026 session 2 application correction begins on jeemain.nta.nic.in; edit details by tomorrow

Latest Question