Question

the sum of n terms of two arithmetic progression are in the ratio (3n+8):(7n+15). find the ratio of their 12th terms?

Answer 1

infinite

Answer 2

WELCOME TO OPEN STUDY!

Answer 3

(2001+1)x1000 + 1001

Answer 4

a = 1
l = 2001
n = 2001

use the arithmetic series formula

Answer 5

Come on, everyone knows this?
1 + 2 +...............+2000 +2001
2001 + 2000 +...........+2 + 1

Answer 6

There are 1000 pairs of numbers that add up to 2001
1000*2001 = 2,001,000

In general, the sum of the consecutive integers from a to b = (a+b)*(b-a+1)/2

Answer 7

use the formula for sum of arithmetic sequence
\[S_n = \frac{n(a_1 + a_{n})}{2}\]

where Sn is the sum
n is the number of terms
a1 is the first term
an is the last term

since the last term is 2001 we can say there are 2001 terms so let's substitute
\[S_n = \frac{2001(1 + 2001)}{2}\]
\[S_n = \frac{2001(2002)}{2}\]
\[S_n = 2001 \times 1001\]
use a calculator now i cannot solve this mentally

Answer 8

1 + 2 + 3 .......................n
The sum to n terms is given by:

\(Sum = \frac{n(n+1)}{2}\)

Use this to find Sum..

n = 2001 here..

Answer 9

ANS: 2,003,001

Answer 10

thanks

Answer 11

Best of luck, Bas.

Answer 12

\[\large \color{green}{\textbf{Welcome To Openstudy..}}\]

Answer 13

Do not edit the question just click on close button and then post the new question..
@basiljohn