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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
The first and last terms of an AP are a and l respectively. Show that the sum of the nth term from the beginning and the nth term form the end is (a + l).
Find an AP whose 4th term is 9 and the sum of its 6th and 13th terms is 40.
First term and the common differences of an A.P. are 6 and 3 respectively; find S27.
Solution: First term = a = 6, common difference = d = 3, S27 = ?
Sn = `"n"/2 [square + ("n" - 1)"d"]` - Formula
Sn = `27/2 [12 + (27 - 1)square]`
= `27/2 xx square`
= 27 × 45
S27 = `square`
In an A.P. the first term is 8, nth term is 33 and the sum to first n terms is 123. Find n and d, the common differences.
If the first term of an A.P. is 2 and common difference is 4, then the sum of its 40 terms is
The 9th term of an A.P. is 449 and 449th term is 9. The term which is equal to zero is
The sum of n terms of two A.P.'s are in the ratio 5n + 9 : 9n + 6. Then, the ratio of their 18th term is
The sum of first 14 terms of an A.P. is 1050 and its 14th term is 140. Find the 20th term.
If the sum of the first four terms of an AP is 40 and that of the first 14 terms is 280. Find the sum of its first n terms.
Find the sum:
`(a - b)/(a + b) + (3a - 2b)/(a + b) + (5a - 3b)/(a + b) +` ... to 11 terms
