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
First Term =
-4
Common Difference =-3
Explanation:
If we use the usual notation and denote the first term of the AP by
u_n = a + (n-1)d
And the sum of the first
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,
(27d)-(-8d) = (-105) - (0)
:. 35d = -105
:. d=-3
Substituting
3a +12= 0 => a=-4
Hence we have:
First Term =
-4
Common Difference =-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