QuestionJuly 15, 2025

Find the first four terms of the sequence given by the following. a_(n)=36-5(n-1),n=1,2,3ldots square ,square ,square ,square

Find the first four terms of the sequence given by the following. a_(n)=36-5(n-1),n=1,2,3ldots square ,square ,square ,square
Find the first four terms of the sequence given by the following.
a_(n)=36-5(n-1),n=1,2,3ldots 
square ,square ,square ,square

Solution
4.1(278 votes)

Answer

36, 31, 26, 21 Explanation 1. Calculate the first term Substitute n=1 into a_n = 36 - 5(n-1) to get a_1 = 36 - 5(1-1) = 36. 2. Calculate the second term Substitute n=2 into a_n = 36 - 5(n-1) to get a_2 = 36 - 5(2-1) = 31. 3. Calculate the third term Substitute n=3 into a_n = 36 - 5(n-1) to get a_3 = 36 - 5(3-1) = 26. 4. Calculate the fourth term Substitute n=4 into a_n = 36 - 5(n-1) to get a_4 = 36 - 5(4-1) = 21.

Explanation

1. Calculate the first term<br /> Substitute $n=1$ into $a_n = 36 - 5(n-1)$ to get $a_1 = 36 - 5(1-1) = 36$.<br />2. Calculate the second term<br /> Substitute $n=2$ into $a_n = 36 - 5(n-1)$ to get $a_2 = 36 - 5(2-1) = 31$.<br />3. Calculate the third term<br /> Substitute $n=3$ into $a_n = 36 - 5(n-1)$ to get $a_3 = 36 - 5(3-1) = 26$.<br />4. Calculate the fourth term<br /> Substitute $n=4$ into $a_n = 36 - 5(n-1)$ to get $a_4 = 36 - 5(4-1) = 21$.
Click to rate: