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प्रश्न
Find the sum of the first 40 positive integers divisible by 3
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उत्तर
In the given problem, we need to find the sum of terms for different arithmetic progressions. So, here we use the following formula for the sum of n terms of an A.P.,
`S_n = n/2 [2a + (n - 1)d]`
Where; a = first term for the given A.P.
d = common difference of the given A.P.
First 40 positive integers divisible by 3
n = number of terms
First 40 positive integers divisible by 3
So, we know that the first multiple of 3 is 3 and the last multiple of 3 is 120.
Also, all these terms will form an A.P. with the common difference of 3.
So here,
First term (a) = 3
Number of terms (n) = 40
Common difference (d) = 3
Now, using the formula for the sum of n terms, we get
`S_n = 40/2 [2(3) + (40 - 1)3]`
= 20[6 + (39)3]
= 20(6 + 117)
= 20(123)
= 2460
Therefore, the sum of first 40 multiples of 3 is 2460
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संबंधित प्रश्न
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Find the sum of the first 25 terms of an A.P. whose nth term is given by an = 7 − 3n
The 8th term of an AP is zero. Prove that its 38th term is triple its 18th term.
If 10 times the 10th term of an AP is equal to 15 times the 15th term, show that its 25th term is zero.
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(Assume the four consecutive terms in A.P. are a – d, a, a + d, a +2d)
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Q.6
Q.14
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 S10 if a = 6 and d = 3
Find the sum:
`4 - 1/n + 4 - 2/n + 4 - 3/n + ...` upto n terms
In an AP, if Sn = n(4n + 1), find the AP.
Assertion (A): a, b, c are in A.P. if and only if 2b = a + c.
Reason (R): The sum of first n odd natural numbers is n2.
