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
The sum of n, 2n, 3n terms of an A.P. are S1 , S2 , S3 respectively. Prove that S3 = 3(S2 – S1 )
In an AP: Given a = 5, d = 3, an = 50, find n and Sn.
The first term of an A.P. is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.
Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.
The first term of an A.P. is 5, the last term is 45 and the sum of its terms is 1000. Find the number of terms and the common difference of the A.P.
The 4th term of an AP is zero. Prove that its 25th term is triple its 11th term.
The 24th term of an AP is twice its 10th term. Show that its 72nd term is 4 times its 15th term.
The first term of an A. P. is 5 and the common difference is 4. Complete the following activity and find the sum of the first 12 terms of the A. P.
a = 5, d = 4, s12 = ?
`s_n = n/2 [ square ]`
`s_12 = 12/2 [10 +square]`
`= 6 × square `
` =square`
The sum of first 14 terms of an A.P. is 1050 and its 14th term is 140. Find the 20th term.
If an = 3 – 4n, show that a1, a2, a3,... form an AP. Also find S20.

