Advertisements
Advertisements
प्रश्न
Read the following passage:
|
India is competitive manufacturing location due to the low cost of manpower and strong technical and engineering capabilities contributing to higher quality production runs. The production of TV sets in a factory increases uniformly by a fixed number every year. It produced 16000 sets in 6th year and 22600 in 9th year. |
- In which year, the production is 29,200 sets?
- Find the production in the 8th year.
OR
Find the production in first 3 years. - Find the difference of the production in 7th year and 4th year.
Advertisements
उत्तर
a6 = 16000, a9 = 22600
a + 5d = 16000 ...(1)
a + 8d = 22600 ...(2)
Substitute a = 1600 – 5d from (1)
16000 – 5d + 8d = 22600
3d = 22600 – 16000
3d = 6600
d = `6600/3` = 2200
a = 16000 – 5(2200)
a = 16000 – 11000
a = 5000
i. an = 29200, a = 5000, d = 2200
an = a + (n – 1)d
29200 = 5000 + (n – 1)2200
29200 – 5000 = 2200n – 2200
24200 + 2200 = 2200n
26400 = 2200n
n = `264/22`
n = 12
In 12th year the production was Rs 29200
ii. n = 8, a = 5000, d = 2200
an = a + (n – 1) d
= 5000 + (8 – 1)2200
= 5000 + 7 × 2200
= 5000 + 15400
= 20400
The production during 8th year is Rs 20400
OR
n = 3, a = 5000, d = 2200
Sn = `n/2[2a + (n - 1)d]`
= `3/2 [2(5000) + (3 - 1) 2200]`
S3 = `3/2 (10000 + 2 xx 2200)`
= `3/2 (10000 + 4400)`
= 3 × 7200
= 21600
The production during first 3 years is 21600
iii. a4 = a + 3d
= 5000 + 3(2200)
= 5000 + 6600
= 11600
a7 = a + 6d
= 5000 + 6 × 2200
= 5000 + 13200
= 18200
a7 – a4 = 18200 – 11600 = 6600
APPEARS IN
संबंधित प्रश्न
If the 3rd and the 9th terms of an AP are 4 and –8 respectively, which term of this AP is zero?
Find the sum of first 22 terms of an AP in which d = 7 and 22nd term is 149.
Find the sum of the first 40 positive integers divisible by 3
Choose the correct alternative answer for the following question .
15, 10, 5,... In this A.P sum of first 10 terms is...
What is the sum of first 10 terms of the A. P. 15,10,5,........?
The first and last term of an A.P. are a and l respectively. If S is the sum of all the terms of the A.P. and the common difference is given by \[\frac{l^2 - a^2}{k - (l + a)}\] , then k =
Which term of the AP 3, 15, 27, 39, ...... will be 120 more than its 21st term?
If a = 6 and d = 10, then find S10
Find the sum of the integers between 100 and 200 that are
- divisible by 9
- not divisible by 9
[Hint (ii) : These numbers will be : Total numbers – Total numbers divisible by 9]
Find the sum of all odd numbers between 351 and 373.

