How to find the first term, the common difference, and the nth term of the arithmetic sequence here? 8th term is 4; 18th term is -96

Answer 1

Theodore Amidon

First term is #74# , common difference is #-10#
#n# th term of A.P. series is #T_n= a +(n-1)d#

Let #a, d ,n# be the first term , common difference and number of

terms of an A.P. series

#n# th term of A.P. series is #T_n= a +(n-1)d#

#8# th term of A.P. series is #T_8= a +(8-1)d=4 #or

#a + 7 d=4 ; (1) #

#18# th term of A.P. series is #T_18= a +(18-1)d=-96 #or

#a + 17 d=-96 ; (2) # , subtracting equation (1) from equation (2)

we get , #10 d= -100 or d = -10 # putting #d=-10# in

equation (1) we get, # a - 70 =4 or a = 74#

Hence first term is #74# and common difference is #-10# [Ans]

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email

Answer 2

Chloe Alexander

The first term is #74# and common difference is #-10#

If a Arithmatic Sequence has first term #a# and the common difference #d# then the formula of #n^(th)# term is #a_n=a+(n-1)d# ......(1)

According to question #8^(th)# term is #4# Put n=8 in equation (1) #=> a_8=a+(8-1)d=a+7d# but #a_8=4# Hence #=>a+7d=4# ........(2) #=> a=4-7d# and #18^(th)# term is #-96# Put n=18 in equation (1) #=> a_18=a+(18-1)d=a+17d# but #a_18=-96# Hence #=>a+17d=-96# ........(3) by putting value of #a# from equation (2) in the equation (3) #=>(4-7d)+17d=-96# #=>4+10d=-96# Transfer #4# to the Right Hand Side #=>10d=-96-4=-100# Divide by 10 #(10d)/10=-100/10# #d=-10#

Put the value of #d# in equation (2) to get the first term of the Arithmetic sequence By Equation (2) #a=4-7d=4-7(-10)# #a=4+70# #a=74#

Hence the first term is #74# and common difference is #-10#

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email

Answer 3

Christian Antunez

Common difference: #d=-10#
first term: #a=74#
nth term: #T_n=74-10d#

8th term #T_8=4#

18th term #T_18=-96#

General formula for nth tern

#T_n=a+(n-1)d#

#(n,T_n)=(8,4)-># #4=a+(8-1)xxd# Simplifying #a+7d=4-----(1)#

#(n,T_n)=(18,-96)-># #-96=a+(18-1)xxd# Simplifying #a+17d=-96---(2)#

Subtrtacting (1) from (2) #10d=-100#

#d=-10# Substituting #d=-10# in (1) #a+7xx(-10)=4# #a-70=4# #a=74#

#T_n=74-10d#

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email

Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.