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
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

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