Advertisements
Advertisements
प्रश्न
A manufacturer of TV sets produces 600 units in the third year and 700 units in the 7th year. Assuming that the production increases uniformly by a fixed number every year, find:
- the production in the first year.
- the production in the 10th year.
- the total production in 7 years.
Advertisements
उत्तर
Since the production increases uniformly by a fixed number every year, the sequence formed by the production in different years in an A.P.
Let the production in the first year = a
Common difference = number of units by which the production increases every year = d
We have,
t3 = 600
`\implies` a + 2d = 600 ...(1)
t7 = 700
`\implies` a + 6d = 700 ...(2)
Subtracting (1) from (2), we get
4d = 100
`\implies` d = 25
`\implies` a + 2 × 25 = 600
`\implies` a = 550
i. The production in the first year = 550 Tv sets
ii. The production in the 10th year = t10
= 550 + 9 × 25
= 775 Tv sets
iii. The total production in 7 years = S7
= `7/2[2 xx 550 + 6 xx 25]`
= `7/2 [1100 + 150]`
= `7/2 xx 1250`
= 4375 Tv sets
संबंधित प्रश्न
The first and the last terms of an AP are 8 and 65 respectively. If the sum of all its terms is 730, find its common difference.
In an AP given a = 8, an = 62, Sn = 210, find n and d.
Find the middle term of the AP 6, 13, 20, ..., 216.
If k, (2k - 1) and (2k + 1) are the three successive terms of an AP, find the value of k.
Find the first term and common difference for the following A.P.:
5, 1, –3, –7, ...
Find whether 0 (zero) is a term of the A.P. 40, 37, 34, 31,... .
Let there be an A.P. with first term 'a', common difference 'd'. If an denotes in nth term and Sn the sum of first n terms, find.
If in an A.P. Sn = n2p and Sm = m2p, where Sr denotes the sum of r terms of the A.P., then Sp is equal to
Q.10
Find the value of a25 – a15 for the AP: 6, 9, 12, 15, ………..
