If the second term of an HP is 1/6 and the fourth terms is 1/14, then find the tenth term of the HP?

Let the required length be L m.
More men, More length (Direct proportion)
Less days, Less length (Direct proportion)
Men 20 : 70
Days 6 : 3
? (20 x 6 x L) = (70 x 3 x 56)
? L = (70 x 3 x 56) / (20 x 6) = 98 m

Here M₁ = 30, D₁ = 16,
T₁ = 9, M₂ = 36 and T₂ = 8
By formula, M₁D₁T₁ = M₂D₂T₂
30 x 16 x 9 = 36 x D₂ x 8
? D₂ = (30 x 16 x 9)/(36 x 8) = 15 days

? For 1000 men, provision lasts for 48 days.
? For 1 men, provision lasts for (48 x 1000) days.
? For (1000 + 600) men, provision will last for (48 x 1000)/1600 days.
? Required number of days = (48 x 1000)/1600 = 3 x 10 = 30

M₁ = 12 H₁ = 16, W₁ = 33600
M₂ = 18, H₂ = 12, W₂ = ?
By formula, M₁H₁/W₁ = M₂H₂/W₂
(12 x 16)/33600 = (18 x 12)/W₂
? W₂ = (33600 x 18)/16
? W₂ = 2100 x 8
? W₂ = 37800

Let the required number of flies be N.
More spiders, More flies (Direct proportion)
More time, More flies (Direct proportion)
? 10 x 10 x N = 200 x 200 x 10
? N = (200 x 200 x 10)/(10 x 10) = 4000 flies

? For 50 students, food is sufficient for 45 days.
? For 1 students, food is sufficient for 45 x 50 days.
? For 75 students, food is sufficient for (45 x 50)/75 days i.e., for 30 days .

Work done = 1/3
Remaining work = (1 - 1/3) = 2/3
Let the number of additional men be N. (Direct proportion)
More days, Less men (Indirect proportion)
Work 1/3 : 2/3
Days 50 : 40
? 1/3 x 50 x (40 + N) = 2/3 x 40 x 40
? 5 x (40 + N) = 2 x 40 x 4
? 200 + 5N = 320
? 5N = 320 - 200 = 120
? N = 120/5 = 24
? Required number of men = 24

Let us assume that each soldier eats one unit of food per day. Thus, total units of food at the beginning will be 1000 x 30 = 30000.
After 10 days 1000 soldiers would have eaten 1000 x 10 = 10000 units of food. Thus food left after 10 days equals 20000 units. Now, there are total of 2000 soldiers who eat one unit of food every day. So, the number of days that 20000 units of food will serve 2000 soldiers is 20000/2000 = 10

Let man's 1 day's work = 1/m
and boy's 1 day's work = 1/n
1 day's work man and boy = 1/24
Man's 6 day's work = 6/m
Now, for 20 days, both man and boy do the work and for last 6 days, only man does the work.
According to the question,
1/m + 1/n = 1/24
? 20(1/m + 1/n) + 6/m = 1
? 20 x (1/24) + 6/m = 1
? 6/m = (1 - 20/24) = 4/24 = 1/6
? 1/n = 1/36
Now from eq. (i)
1/m + 1/n = 1/24
1/36 + 1/n = 1/24
? 1/n = (1/24) - (1/36) = 1/72
Hence, the boy alone can do the work in 72 days.

12 men ? 18 women
? 1 man ? 18/12 women ? 3/2 women
? 8 men ? (3/2) x 8 = 12 women
? 8 men + 16 women = 12 women + 16 women = 28 women
? 18 women can do the work in 14 days .
? 1 woman can do the same work in (14 x 18) days.
? 28 women will do the same work in (14 x 18)/28 days.
? Required number of days = (14 x 18)/28 = 9 days