Find the sum of 10th terms of the PRIME integers, where , whose nth term is&nb

Find the sum of 10th terms of the PRIME integers, where $\mathrm{n= a^{2}}$ , whose nth term is $\mathrm{2\left ( n-1 \right )\times n}$ .
Option: 1
320

Option: 2
660

Option: 3
440

Option: 4
820

Answers (1)

To find the sum of the 10th terms of the prime integers, where $\mathrm{n= a^{2}}$ and a is a positive integer, with the nth term as $\mathrm{2\left ( n-1 \right )\times n}$ , we need to calculate the value of each term and then add them up.

First, let's determine the 10th prime number:
2, 3, 5, 7, 11, 13, 17, 19, 23, 29

Now, let's substitute the values of $\mathrm{n= a^{2}}$ into the given expression for each term:

For the first term (n = 1):
$\mathrm{2(1-1) \times 1=2 \times 0 \times 1=0}$

For the second term (n=2) :
$\mathrm{2(2-1) \times 2=2 \times 1 \times 2=4}$

For the third term (n=3) :
$\mathrm{2(3-1) \times 3=2 \times 2 \times 3=12}$

For the fourth term (n=4) :
$\mathrm{2(4-1) \times 4=2 \times 3 \times 4=24}$

For the fifth term (n=5) :
$\mathrm{2(5-1) \times 5=2 \times 4 \times 5=40}$

For the sixth term (n=6) :
$\mathrm{2(6-1) \times 6=2 \times 5 \times 6=60}$

For the seventh term (n=7) :
$\mathrm{2(7-1) \times 7=2 \times 6 \times 7=84}$

For the eighth term (n=8) :
$\mathrm{c 2(8-1) \times 8=2 \times 7 \times 8=112}$

For the ninth term (n=9) :
$\mathrm{2(9-1) \times 9=2 \times 8 \times 9=144}$

For the tenth term (n=10) :
$\mathrm{2(10-1) \times 10=2 \times 9 \times 10=180}$
Now, let's find the sum of the 10th terms:

$\mathrm{\text { Sum }=0+4+12+24+40+60+84+112+144+180=660}$

Therefore, the sum of the 10th terms of the prime integers, where $\mathrm{n=a^2}$ and a is a positive integer, with the n th term as $\mathrm{2(n-1) \times n}$ , is 660 .