NECO GCE MATHS ANSWERS

(7ai)
T₃ = -15
T₁₁ = 25
Tn = a+(n-1)
a+(3-1)d= -15
a+2d= -15 ….(i)
a+(11-1)d=25
a+10d=25 ….(ii)
Substract equatiom (i) from (ii)
a+2d = -15
– a+10d= 25
———————
8d = 40
d = 40/8
d = 5
Common difference (d)= 5

(7aii)
Substitute d in equ(i)
a+2d= -15
a+2(5)= -15
a+10= -15
a= -15-10
a= -25
First term (a)= -25

(7aiii)
Sₙ= ⁿ/₂ (2a+(n-1)d)
30=ⁿ/₂ (2x -25 + (n-1) 5)
30×2= n(-150 + 5n – 5)
60= -150n + 5n² – 5n
60= -155n + 5n²
5n² – 155n – 60 =0
n² – 11n – 12=0
n² – 12n + n – 12=0
n(n-12) +1(n-12)=0
(n-12)(n+1)=0
n-12=0 or n+1=0
n=12 or n=-1
Number of terms=12

(7b)
x = k/y²
2=k/4²
2×4² = k
k=32
x=32/y²
8=32/y²
8y²=32
y²=32/8
y²=4
y=±√4
y=±2
y=2