Consider the length of cable as the term of series.
From 1st cable to 11th cable, the length increases at a common difference of 1.75m,
The longest cable as per question is 11th cable,
First summarizing the question,
1st term(A)=3.5m
common difference (d)=1.75m
11th term(t11)=?
Now,
tn=A+(n-1)×d
or,t11=3.5+(11-1)×1.75
=3.5+10×1.75
=3.5+17.5=21 metre ANS1
Now,the question is how many of first few cables can be cut out before the first spool runs out .Now, we need to understand that the cables are cut out from the spooler one by one and designed as explained in question .Since
only 100m cable in first spool ,we need to find in how many times of cabling the 100 m is just about to finish so that leftover can be cut out.
First find the sum up to which the length is less or equal to 100m
Sn ≤ 100 metre
or,n/2×[2×3.5+(n-1)×1.75] ≤ 100
Put n=9 we get Sn=94.5 which is ≤ 100
Put n=10 we get Sn=113.75 which is >100
So, when the cable is cut till 9th cable, a few cable is leftover which can be cut out
Cable that can be cut out before 1st spool runs out=100-94.5=5.5m ans2
After 11th cable , the cables are kept in order of geometric series ,
If t12=x then t13=55%×x
common ratio(r)=t13/t12=55%=0.55
Now to find the tn term which is greater or equal to 1 metre,
tn ≥ 1
or, a(r)^(n-1) ≥ 1
or(55%×t11)×{(0.55)^(n-1)} ≥1
or(0.55×21)×{(0.55)^(n-1)} ≥1
or,21×{0.55^(n-1+1)} ≥ 1
or,21×0.55^n ≥ 1
Put n=5 then t5=1.056 ≥1
Put n=6 then t6=0.5812<1
So no of cables in 2nd part of bridge=5 ans3
From 1st part S11==X=11/2×[3.5+21]=134.75
From 2nd part ,
Sum of 5 cables in G.P. IS
S5=Y
=a(1-r^5)/(1-r)
=11.55×(1-0.55^5)/(1-0.55)
=24.37 metre
Therefore total amount of cabling required for both parts of bridge
=X+Y
= 134.75+24.37
=159.12 metre ans4
