00:01
In the first question here, we're given an ap, where the third term is four times the sixth term, and the sum of the first 10 terms is 150.
00:12
Now, in general, for ap, a term, the n -term will be a plus n -1 -d, where a is the first term, d is the common difference, and n is the anthem.
00:27
And sum to n -term will be n -over -2, 2 -a -plus.
00:33
And minus 1d.
00:39
I want to find the first term and a common difference.
00:42
So i'm going to let a be the first term and d be the common difference.
00:47
So we are given the term is four times the sixth term.
00:51
So that is t3 equals 4 of t6.
00:55
So that will be t3 is a plus 3 minus 1 d.
01:02
Because to four times of t6 will be a plus 6 minus 1 d.
01:07
So we have a plus 2d is 4 of a plus 5d.
01:18
Spending out on the right, we get this.
01:24
So i'm going to bring d to the left and a to the right.
01:27
I have minus 18d is 3a.
01:30
So a is minus 6d.
01:35
Now we're also given that the sum of the first 10 terms is 150.
01:40
So we have s10 is equals to 150.
01:45
So that will be 10 over 2, 2a plus n is 10, 10 minus 1d is equal to 150.
01:59
So this is 5, bring over to the right side is 150, divide by 5.
02:04
I'll get 30.
02:07
On the left, i have 2a plus 90.
02:10
I'm going to sub the a with minus 60.
02:20
So i get d is minus 10.
02:24
And my a is minus 6d, which is minus 10, and so my a is 60.
02:33
So therefore, first term is equals to a is 60, and common difference is d, and that will be minus 10.
02:52
In part b, we want to find the least number of terms of the ap such that the sum is negative...