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  • Determine the total sum of the 10th terms of the PRIME integers, where the nth term is expressed as <img alt="\mathrm{5 n^2-3 n+1.}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7

Determine the total sum of the 10th terms of the PRIME integers, where the nth term is expressed as \mathrm{5 n^2-3 n+1.}

Option: 1

3100


Option: 2

5002


Option: 3

2370


Option: 4

6010


Answers (1)

best_answer

To find the sum of the 10 th terms of the sequence generated by the expression \mathrm{5 n^2-3 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{ \operatorname{Sum}=\left(5(1)^2-3(1)+1\right)+\left(5(2)^2-3(2)+1\right)+\ldots+\left(5(10)^2-3(10)+1\right) }
Simplifying the expression for each term, we have:

\mathrm{ \text { Sum }=(5-3+1)+(20-6+1)+\ldots+(500-30+1) }
Simplifying further, we get:

\mathrm{ \text { Sum }=3+15+\ldots+471 }

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 3 , the last term is 471 , and the number of terms is 10 .

Plugging these values into the formula, we get:

\mathrm{ \text { Sum }=(10 / 2)(3+471)=5(474)=2370 }
Therefore, the sum of the \mathrm{10^{\text {th }}} terms of the sequence generated by \mathrm{ 5 n^2-3 n+1 \, \, is\, \, 2370 .}

Posted by

SANGALDEEP SINGH

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