How do you write a rule for the nth term of the arithmetic sequence and then find #a_10# for #d=5, a_5=33#?

Question

Lily Anderson · Accepted Answer

#a_n = 13 + 5*(n-1)#
#a_10 = 58# The general form of #n^(th)# term of an AP is given as:- #a_n = a_1 + (n-1)*d# In this case, #d = 5 # #therefore a_n = a_1 + 5*(n-1)# #a_5 = a_1 + 5*(5-1) = a_1+5*4 = a_1 + 20# But, #a_5 = 33# #therefore 33 = a_1 +20 # #=> a_1 = 33-20 = 13# Hence #a_n = 13 + 5*(n-1)# #a_10 = 13 + 5*(10-1) = 13+5*9 = 13+45 = 58#

Caleb Alexander · Answer

undefined To write a rule for the nth term of an arithmetic sequence, you can use the formula: $ a_n = a_1 + (n - 1) \times d $, where $ a_n $ is the nth term, $ a_1 $ is the first term, $ n $ is the term number, and $ d $ is the common difference.

Given $ a_5 = 33 $ and $ d = 5 $, first find $ a_1 $ by substituting the values into the formula. Then, use the same formula to find $ a_{10} $ by substituting $ n = 10 $ and the found value of $ a_1 $.

Step 1: Substitute $ a_5 = 33 $ and $ d = 5 $ into the formula to find $ a_1 $:
$$ a_5 = a_1 + (5 - 1) \times 5 $$
$$ 33 = a_1 + 4 \times 5 $$
$$ 33 = a_1 + 20 $$
$$ a_1 = 33 - 20 $$
$$ a_1 = 13 $$

Step 2: Use the formula to find $ a_{10} $ with $ n = 10 $:
$$ a_{10} = 13 + (10 - 1) \times 5 $$
$$ a_{10} = 13 + 9 \times 5 $$
$$ a_{10} = 13 + 45 $$
$$ a_{10} = 58 $$

Therefore, $ a_{10} = 58 $ for the given arithmetic sequence.