In an arithmetric progression, the fifth term is four times the first term and the sum of the first 10 terms is -175. Look for first term and common difference ?

2 Answers
Steve M
Jan 4, 2018

First Term = #-4#
Common Difference = #-3#

Explanation:

If we use the usual notation and denote the first term of the AP by #a# and the common difference by #d#, then, the #n^(th)# term is:

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

And the sum of the first #n# term is:

# S_n = n/2{2a+(n-1)d} #

And so we can form the equations:

# u_5 = 4u_1 => a+4d = 4a #
# :. 3a-4d = 0 # ..... [A]

And:

# S_10=-175 => 5(2a+9d)=-175 #
# :. 2a+9d = -35 # .... [B]

And we now solve [A] and [B] simultaneously, #3Eq[B]-2Eq[A]# gives:

# (27d)-(-8d) = (-105) - (0) #
# :. 35d = -105 #
# :. d=-3 #

Substituting #d# into #Eq[A]#

# 3a +12= 0 => a=-4 #

Hence we have:

First Term = #-4#
Common Difference = #-3#

Jim G.
Jan 4, 2018

#a_1=-4" and "d=-3#

Explanation:

#"using the n th term and sum to n terms formulae"#
#"for an arithmetic progression"#

#•color(white)(x)a_n=a+(n-1)d#

#•color(white)(x)S_n=n/2[2a_1+(n-1)d]#

#rArrS_(10)=5(2a_1+9d)=-175#

#rArr10a_1+45d=-175#

#rArr10a_1=-175-45d#

#rArra_1=-17.5-4.5d to(1)#

#"now "a_5=4a_1=a_1+4d#

#rArr3a_1=4drArra_1=(4d)/3to(2)#

#rArr3(-17.5-4.5d)=4dlarr"substituting "(1)#

#rArr-52.5-13.5d=4d#

#rArr17.5d=-52.5rArrd=-3#

#"substituting in "(2)#

#a_1=(4xx-3)/3=-4#

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