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प्रश्न
Find the sum of two middle most terms of the AP `-4/3, -1 (-2)/3,..., 4 1/3.`
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उत्तर
Here, first term (a) = `-4/3`,
Common difference (d) = `-1 + 4/3 = 1/3`
And the last term (l) = `4 1/3 = 13/3` ...[∵ nth term of an AP, l = an = a + (n – 1)d]
⇒ `13/3 = -4/3 + (n - 1)1/3`
⇒ 13 = – 4 + (n – 1)
⇒ n – 1 = 17
⇒ n = 18 ...[Even]
So, the two middle most terms are `(n/12)^("th")` and `(n/2 + 1)^("th")`
i.e., `(18/n)^("th")` and `(18/2 + 1)^("th")`terms
i.e., 9th and 10th terms.
∴ a9 = a + 8d
= `- 4/3 + 8(1/3)`
= `(8 - 4)/3`
= `4/3`
And a10 = `-4/3 + 9(1/3)`
= `(9 - 4)/3`
= `5/3`
So, sum of the two middle most terms
= a9 + a10
= `4/3 + 5/3`
= `9/3`
= 3
संबंधित प्रश्न
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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,
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Therefore, the sum of natural numbers between 1 to 140, which are divisible by 4 is `square`.
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