Math, asked by tanmaybhere100, 1 year ago

A mobile company manufactured 2000 mobile in the fifth month and 2750 mobiles in the 8th month. Assuming that the production increases informally by fixed number every month find the production of mobiles in the first month and in the 10th month.

Answers

Answered by jainsk03
13

Answer:

Step-by-step explanation:

Units in 5th month=2000

Units in 8th month=2750

Therefore in 3 months increase is 750

Therefore increase in one month is 750/3=250

Now in fifth month units are 2000

Therefore in first month they will be 2000-4(250) ...(4 because there are four months between 1st and 5th month)

2000-1000=1000units

In 8 months there are 2750units

Therefore in 10 Montana there will be

2750+2(250)=2750+500=3250 units


tanmaybhere100: Nice
jainsk03: Thanks bro. Please mark as the brainliest
vishu8730: Was it really helpful
vishu8730: did you understand it
tanmaybhere100: Ohh srry I made to him
vishu8730: its ok
jainsk03: No problem bro. Actually his method is very useful in competitive exams
tanmaybhere100: Ok bye
Answered by Harshitm077
3

Answer:

Production of mobiles in first month = 1000 units and in 10th month = 3250 units.

Step-by-step explanation:

Mobiles manufactured in 5th month are 2000 and in 8th month are 2750.

Since production increases by fixed number in every month then it can be considered as an Arithmetic Progression (AP).

Let a be the first term and d be the common difference.

Now, according to the formula for the nth term,

aₙ = a + (n - 1)d

a₅ = a + 4d = 2000 .......(i)

a₈ = a + 7d = 2750 ........(ii)

Solving the equations (i) and (ii) for a and d we get,

a = 1000 and d = 250

Now, a₁ = a = 1000

         a₁₀ = a + 9d = 1000 + 9 × 250

         a₁₀ = 3250

Thus, first month a₁ = 1000 and,

         10th month a₁₀ = 3250.

Similar questions