Advertisements
Advertisements
प्रश्न
A sum of Rs. 700 is to be paid to give seven cash prizes to the students of a school for their overall academic performance. If the cost of each prize is Rs. 20 less than its preceding prize; find the value of each of the prizes.
उत्तर
Total amount of Prize Sn = Rs. 700
Let the value of the first prize be Rs. a
Number of prizes = n = 7
Let the value of first prize be Rs. a
Depreciation in next prize = Rs. 20
We have,
`S_n = n/2 [2a + (n - 1)d]`
`\implies 700 = 7/2 [2a + 6(-20)]`
`\implies 700 = 7/2 [2a = 120]`
`\implies` 1400 = 14a – 840
`\implies` 14a = 2240
`\implies` a = 160
`\implies` Value of 1st Prize = Rs. 160
Value of 2nd Prize = Rs. (160 – 20) Rs. 140
Value of 3rd Prize = Rs. (140 – 20) = Rs. 120
Value of 4th Prize = Rs. (120 – 20) = Rs. 100
Value of 5th Prize = Rs. (100 – 20) Rs. 80
Value of 6th Prize = Rs. (80 – 20) = Rs. 60
Value of 7th Prize = Rs. (60 – 20) = Rs. 40
APPEARS IN
संबंधित प्रश्न
Find the sum given below:
34 + 32 + 30 + ... + 10
In an AP given a = 8, an = 62, Sn = 210, find n and d.
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.
If the 10th term of an AP is 52 and 17th term is 20 more than its 13th term, find the AP
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`
If 18th and 11th term of an A.P. are in the ratio 3 : 2, then its 21st and 5th terms are in the ratio
Let the four terms of the AP be a − 3d, a − d, a + d and a + 3d. find A.P.
Find the sum of natural numbers between 1 to 140, which are divisible by 4.
Activity: Natural numbers between 1 to 140 divisible by 4 are, 4, 8, 12, 16,......, 136
Here d = 4, therefore this sequence is an A.P.
a = 4, d = 4, tn = 136, Sn = ?
tn = a + (n – 1)d
`square` = 4 + (n – 1) × 4
`square` = (n – 1) × 4
n = `square`
Now,
Sn = `"n"/2["a" + "t"_"n"]`
Sn = 17 × `square`
Sn = `square`
Therefore, the sum of natural numbers between 1 to 140, which are divisible by 4 is `square`.
An AP consists of 37 terms. The sum of the three middle most terms is 225 and the sum of the last three is 429. Find the AP.