Question

11. A TV manufacturer has produced 1000 TVs in the seventh year and 1450 TVs in the
tenth year. Assuming that the production increases uniformly by a fixed number every
year, find the number of TVs produced in the first year and in the 15th year.

Answer 1

AP ( Arithmetic progression).
An arithmetic progression is a list of numbers a1, a2, a3 ………….. an in which each term is obtained by adding a fixed number to the preceding term except the first term.
This fixed number is called the common difference( d ) of the AP. Common difference of an AP will be the difference between any two consecutive terms.

a2= a1+d
a3= a2+d
a4= a3+d
……..
an= an-1+d ………

Each of the numbers in the list is called a term .

Method to find the common difference :
d = a2 - a1 or a3 - a2 or a4 - a3...

General form of an AP.:
a, a+d, a+2d, a+3d…….

Here a is the first term and d is common difference.

General term or nth term of A.P
The general term or nth term of A.P is given by an = a + (n – 1)d, where a is the first term, d is the common difference and n is the number of term.

SOLUTION :
Given : Number of TVs produced in the seventh year = 1000
t7 = 1000

Number of TVs produced in the tenth year = 1450
t10 = 1450

tn = a + (n – 1)d
t7 = 1000
Here, n = 7
1000 = a + (7-1)d
a + 6 d = 1000 ………...(1)

t10 = 1450
Here, n = 10
1450 = a + (10 - 1) d
a + 9 d = 1450 ………..(2)

On Subtracting equation 2 from equation 1
a + 6 d = 1000
a + 9 d = 1450
(-) (-) (-)
------------------
-3d = -450
d = -450/(-3)
d = 150

Put d = 150 in  equation 1
a + 6 d = 1000
a + 6(150) = 1000
a + 900 = 1000
a = 1000 -900
a = 100

Hence,the number of TVs produced on the Ist year is 100.

We have to find the 15th term of the A.P to find number of TVs produced in the 15th year.
tn = a + (n-1) d
Here, a = 100, n = 15 , d = 150
t15 = 100 + (15-1) 150
t15 = 100 + 14(150)
t15= 100 + 2100
t15= 2200

Hence, the number of TVs produced in the first year is 100 and in the 15th year is 2200.

HOPE THIS WILL HELP YOU….

Answer 2

It is given the production of T.V's

imcreases uniformly .

Let each year production = d

This is in Arithmetic Progression .

Let a , d are first year production

and each year increase in production.

************************************

We know that ,

nth term in A.P = an

an = a + ( n - 1 )d

*******************************†*******

a7 = 1000

=> a + 6d = 1000 ---( 1 )

a10 = 1450

a + 9d = 1450 ---( 2 )

Subtract ( 1 ) from ( 2 ) ,we get

3d = 450

=> d = 450/3

d = 150 ---( 3 )

Substitute d = 150 in equation ( 1 ),

a + 6 × 150 = 1000

=> a + 900 = 1000

a = 1000 - 900

a = 100,

Now ,

Production of T.V's in end of the

15th year ,

a = 100 , d = 150 , n = 15

an = a + ( n - 1 )d

=> a15 = 100 + ( 15 - 1 )×150

= 100 + 14 × 150

= 100 + 2100

= 2500

Therefore ,

First year production = a = 100

15th year production = a15 = 2500

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