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प्रश्न
Find the sum of the following arithmetic progressions:
41, 36, 31, ... to 12 terms
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.
n = number of terms
41, 36, 31, ... to 12 terms
Common difference of the A.P. (d) = `a_2 - a_1`
= 36 - 41
= -5
Number of terms (n) = 12
The first term for the given A.P. (a) = 41
So, using the formula we get,
`S_12 = 12/2 [2(41) + (12 - 1)(-5)]`
= (6)[82 + (11)(-5)]
= (6)[82 - 55]
= (6)[27]
= 162
Therefore the sum of first 12 terms for the given A.P is 162
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संबंधित प्रश्न
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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"]`
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