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प्रश्न
Find the sum given below:
–5 + (–8) + (–11) + ... + (–230)
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उत्तर
–5 + (–8) + (–11) + ... + (–230)
For this A.P.,
a = −5
l = −230
d = a2 − a1
= (−8) − (−5)
= − 8 + 5
= −3
Let −230 be the nth term of this A.P.
l = a + (n − 1)d
−230 = − 5 + (n − 1) (−3)
−225 = (n − 1) (−3)
(n − 1) = 75
n = 76
And Sn = `n/2(a+1)`
= `76/2[(-5)+(-230)]`
= 38 × (-235)
= -8930
संबंधित प्रश्न
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Which term of the AP 21, 18, 15, … is zero?
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Q.2
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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"]`
Sn = 17 × `square`
Sn = `square`
Therefore, the sum of natural numbers between 1 to 140, which are divisible by 4 is `square`.
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