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प्रश्न
Find the sum of first 20 terms of the following A.P. : 1, 4, 7, 10, ........
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उत्तर
Let a be the first term and d be the common difference of the given A.P. Then, we have a = 1 and d = 3.
We have to find the sum of 20 terms of the given A.P.
Putting a = 1, d = 3, n = 20 in
`S_n = \frac{n}{ 2 } [2a + (n - 1) d]`
∴ `S_20 = \frac {20}{2} [2 × 1 + (20 - 1) × 3]`
= 10 (2 + 57)
= 10 × 59
= 590
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संबंधित प्रश्न
If the 3rd and the 9th terms of an AP are 4 and –8 respectively, which term of this AP is zero?
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Find the sum of the first 40 positive integers divisible by 5
Find the sum of all 3-digit natural numbers, which are multiples of 11.
Write the common difference of an A.P. whose nth term is an = 3n + 7.
If the sum of three consecutive terms of an increasing A.P. is 51 and the product of the first and third of these terms is 273, then the third term is
The 9th term of an A.P. is 449 and 449th term is 9. The term which is equal to zero is
Q.2
For an A.P., If t1 = 1 and tn = 149 then find Sn.
Activitry :- Here t1= 1, tn = 149, Sn = ?
Sn = `"n"/2 (square + square)`
= `"n"/2 xx square`
= `square` n, where n = 75
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`.
