Answer:
The correct answer is = 15.
Step-by-step explanation:
Formula:
The sum of the first n terms of an arithmetic progression with first term a and constant difference d is
[tex]S_n=\dfrac{n}{2}[2a+(n-1)d[/tex]
using this formula in this problem
Solution:
The sum of the first ten terms is
[tex]S_{10}=\dfrac{10}{2}[2a+(10-1)d[/tex]
[tex]S_{10}=5(2a+9d)[/tex]
The sum of the 20th, 21st, and 22nd terms is three times the 21st term:
[tex]3a_{21}=3(a+(21-1)d)[/tex]
[tex]3a_{21}=3(a+20d)[/tex]
[tex]3a_{21}=3a+60d[/tex]
The problem then tells us
[tex]S_{10}=3a_{21}[/tex]
[tex]10a+45d=3a+60d[/tex]
[tex]7a=15d[/tex]
there are only positive integers and the first term a is less than 20 as given. Since 7 and 15 have no common factor, the only explanation of the requirements is a = 15 and d = 7. So the progression is
then, 15, 22, 29, 36, ...
The problem says to find the number of terms n for which the sum is 960:
putting value in the formula
[tex]30n+7n^{2}-7n=1920\\7n^{2}+23n-1920=0[/tex]
solving quadratic will give n = 15
thus, the correct answer is 15.
