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प्रश्न
Find the sum of 28 terms of an A.P. whose nth term is 8n – 5.
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उत्तर
nth term of an A.P. = tn = 8n – 5
Let a be the first term and d be the common difference of this A.P.
Then,
a = t1
= 8 × 1 – 5
= 8 – 5
= 3
t2 = 8 × 2 – 5
= 16 – 5
= 11
∴ d = t2 – t1
= 11 – 3
= 8
The sum of n terms of an A.P. = `S = n/2 [2a + (n - 1)d]`
`=>` Sum of 28 terms of an A.P. = `28/2 [2 xx 3 + 27 xx 8]`
= 14[6 + 216]
= 14 × 222
= 3108
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संबंधित प्रश्न
The sum of n terms of three arithmetical progression are S1 , S2 and S3 . The first term of each is unity and the common differences are 1, 2 and 3 respectively. Prove that S1 + S3 = 2S2
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Show that (a – b)2, (a2 + b2) and (a2 + b2) are in AP.
The first three terms of an AP are respectively (3y – 1), (3y + 5) and (5y + 1), find the value of y .
Write the expression for the common difference of an A.P. whose first term is a and nth term is b.
If the sum of n terms of an A.P. is 2n2 + 5n, then its nth term is
If 18th and 11th term of an A.P. are in the ratio 3 : 2, then its 21st and 5th terms are in the ratio
The common difference of the A.P. \[\frac{1}{2b}, \frac{1 - 6b}{2b}, \frac{1 - 12b}{2b}, . . .\] is
Two cars start together in the same direction from the same place. The first car goes at uniform speed of 10 km h–1. The second car goes at a speed of 8 km h–1 in the first hour and thereafter increasing the speed by 0.5 km h–1 each succeeding hour. After how many hours will the two cars meet?
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`.
