Exercise 5.4

Board-exam working:

Here a=121 and d=117−121=−4.

Thus aₙ=121+(n−1)(−4)=125−4n.

For aₙ<0, 125−4n<0, so 125<4n and n>31.25.

The smallest positive integer satisfying this is n=32.

Check the boundary: a₃₁=125−4×31=1, while a₃₂=125−4×32=−3.

Therefore the 32nd term is the first negative term.

Board-exam working:

Let a₃=a+2d and a₇=a+6d.

Their sum is 2a+8d=6, so a+4d=3.

Therefore a₃=3−2d and a₇=3+2d.

Their product is 8: (3−2d)(3+2d)=8.

Thus 9−4d²=8, so 4d²=1 and d=±1/2.

Since a=3−4d, for d=1/2 we get a=1

for d=−1/2 we get a=5.

S₁₆=16/2[2a+15d]=8[2(3−4d)+15d]=8(6+7d)=48+56d.

Hence S₁₆=76 when d=1/2, or S₁₆=20 when d=−1/2.

Board-exam working:

Convert 2.5 m to 250 cm.

With 25 cm between consecutive rungs, the number of gaps is 250÷25=10.

Hence the number of rungs is 10+1=11.

The rung lengths decrease uniformly from 45 cm to 25 cm, so they form an AP with a=45,l=25,n=11.

Total wood length=S₁₁=11/2(45+25)=11/2×70=11×35=385 cm=3.85 m.

Board-exam working:

The sum before house x is 1+2+…+(x−1)=x(x−1)/2.

The total from 1 to 49 is 49×50/2=1225.

The sum after x is 1225−x(x+1)/2.

Equate the two sums: x(x−1)/2=1225−x(x+1)/2.

Multiplying by 2 gives x(x−1)+x(x+1)=2450, so 2x²=2450 and x²=1225.

Thus x=±35, but a house number must be positive, so x=35.

Check: 1+…+34=595 and 36+…+49=595.

Board-exam working:

The first step layer has cross-sectional area (1/2)(1/4), the second contributes twice this amount, and so on up to 15 layers.

With length 50 m, total volume is 50×1/2×1/4(1+2+…+15).

The sum 1+2+…+15 is 15/2(1+15)=15/2×16=120.

Therefore volume=50×1/2×1/4×120=25/4×120=25×30=750 m³.