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Mathematics
UTME / JAMB 1978
Algebra
Solution
An arithmetic progression has first term 11 and fourth term 32. The sum of the first nine terms is
A
351
B
531
C
135
D
153
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
1st term a = 11, 4th term = 32
nth term = a + (n - 1)d
4th term = 11 = (4 - 1)d
= 11 + 3d
= 32
3d = 21
d = 7
sn = \(\frac{n}{2}\)(2a + (n - 1)d)
sn = \(\frac{9}{2}\)(2 \(\times\) 11) + (9 - 1)7
\(\frac{9}{2}\)(22 + 56) = \(\frac{9}{2}\) x 78
= 351
Source: Myschool Past Questions Archive (Record #32708)
Mathematics
UTME / JAMB 1978
Algebra
Solution
The set of value of x and y which satisfies the equations x2 - y - 1 = 0 and y - 2x + 2 = 0 is
A
1, 0
B
1, 1
C
2, 2
D
0, 2
E
1, 2
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
x2 - y - 1 = 0.......(i)
y - 2x + 2 = 0......(ii)
By re-arranging eqn. (ii)
y = 2x - 2........(iii)
Subst. eqn. (iii) in eqn (i)
x2 - (2x - 2) - 1 = 0
x2 - 2x + 1 = 0
= (x - 1) = 0
When x - 1 = 0
x = 1
Sub. for x = 1 in eqn. (iii)
y = 2 - 2 = 0
x = 1, y = 0
Source: Myschool Past Questions Archive (Record #32714)
Mathematics
UTME / JAMB 1978
Algebra
Solution
If x\(^4\) - kx\(^3\) + 10x\(^2\) + lx - 3 is divisible by (x - 1), and if when it is divided by (x + 2) the remainder is 27, find the constants k and l
A
k = -7, 1 = -15
B
k = -15, 1 = -7
C
k = \(\frac{15}{3}\) , 1 = -7
D
k = \(\frac{7}{3}\) , 1 = -17
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
If k = -7 is put as l = -15, the equation x\(^4\) - kx\(^3\) + 10x\(^2\) + lx - 3 becomes x\(^4\) - (7x\(^3\)) + 10x\(^2\) + (15) -3 = x\(^4\) + 7x\(^3\) + 10x\(^2\) - 15x - 3
This equation is divisible by (x - 1) and (x + 2) with the remainder as 27
k = -7, l = -15
Source: Myschool Past Questions Archive (Record #32723)