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प्रश्न
Write the next two terms of the A.P.: `sqrt(27), sqrt(48), sqrt(75)`......
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उत्तर
Given A.P. is `sqrt(27), sqrt(48), sqrt(75)` ......
Here, a1 = `sqrt(27) = 3sqrt(3)`
a2 = `sqrt(48) = 4sqrt(3)`
∴ Common difference = `4sqrt(3) - 3sqrt(3)`
= `sqrt(3) (4 - 3) = sqrt(3)`
Now, Given a3 = `sqrt(75) = 5sqrt(3)`
∴ a4 = `6sqrt(3) = sqrt(108)`
And a5 = `7sqrt(3) = sqrt(147)`
Hence, next two terms are `sqrt(108)` and `sqrt(147)`.
संबंधित प्रश्न
If the 9th term of an A.P. is zero, then prove that 29th term is double of 19th term.
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34 + 32 + 30 +...+10
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(-5)+(-8)+(-11)+...+(-230)
Given Arithmetic Progression 12, 16, 20, 24, . . . Find the 24th term of this progression.
Select the correct alternative and write it.
What is the sum of first n natural numbers ?
If the sum of first n terms of an AP is n2, then find its 10th term.
How many two-digit numbers are divisible by 5?
Activity :- Two-digit numbers divisible by 5 are, 10, 15, 20, ......, 95.
Here, d = 5, therefore this sequence is an A.P.
Here, a = 10, d = 5, tn = 95, n = ?
tn = a + (n – 1) `square`
`square` = 10 + (n – 1) × 5
`square` = (n – 1) × 5
`square` = (n – 1)
Therefore n = `square`
There are `square` two-digit numbers divisible by 5.
Decide whether 301 is term of given sequence 5, 11, 17, 23,.....
Activity :- Here, d = `square`, therefore this sequence is an A.P.
a = 5, d = `square`
Let nth term of this A.P. be 301.
tn = a + (n – 1) `square`
301 = 5 + (n – 1) × `square`
301 = 6n – 1
n = `302/6` = `square`
But n is not positive integer.
Therefore, 301 is `square` the term of sequence 5, 11, 17, 23,......
If p - 1, p + 3, 3p - 1 are in AP, then p is equal to ______.
Find a and b so that the numbers a, 7, b, 23 are in A.P.
