Advertisements
Advertisements
प्रश्न
The sum of three numbers in AP is 3 and their product is -35. Find the numbers.
उत्तर
Let the required numbers be (a - d ),a and (a +d).
Then (a-d) +a +(a+d) = 3
⇒ 3a = 3
⇒ a = 1
Also , (a- d) .a . (a +d) = -35
⇒ a,`(a^2 - d^2 ) = -35`
⇒ 1.`(1-d^2 ) =-35`
⇒ `d^2 = 36 `
⇒ `d = +- 6`
Thus , a = 1 and d = `+-6`
Hence, the required numbers are (-5 ,1 and 7 ) or (7,1 and -5)
APPEARS IN
संबंधित प्रश्न
If the sum of first 7 terms of an A.P. is 49 and that of its first 17 terms is 289, find the sum of first n terms of the A.P.
If the mth term of an A.P. is 1/n and the nth term is 1/m, show that the sum of mn terms is (mn + 1)
Find the sum of first 8 multiples of 3
How many terms of the A.P. 63, 60, 57, ... must be taken so that their sum is 693?
Find the sum of two middle most terms of the AP `-4/3, -1 (-2)/3,..., 4 1/3.`
The first and the last terms of an A.P. are 7 and 49 respectively. If sum of all its terms is 420, find its common difference.
Solve for x : 1 + 4 + 7 + 10 + … + x = 287.
How many terms of the A.P. 24, 21, 18, … must be taken so that the sum is 78? Explain the double answer.
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`.
Find the sum of first 20 terms of an A.P. whose nth term is given as an = 5 – 2n.
