Advertisements
Advertisements
प्रश्न
Find the sum of the following arithmetic series:
(-5)+(-8)+(-11)+...+(-230)
Advertisements
उत्तर
The given arithmetic series is (-5)+(-8)+(-11)+...+(-230)
Here , a = -5 , d = -8 -(-5) = -8 + 5= -3 and l = 230
Let the given series contain n terms. Then,
an = -230
⇒ -5 + (n-1) × (-3) = -230 [ an = a + (n-1) d]
⇒ - 3n - 2 = -230
⇒ -3n = -230 + 2 = -228
⇒ n = 76
`∴ "Required sum" = 76/2 xx [ ( -5) + (-230) ] [ s_n = n/2 (a+l)]`
`= 76/2 xx (-235) `
= - 8930
APPEARS IN
संबंधित प्रश्न
Find the term t15 of an A.P. : 4, 9, 14, …………..
Find the sum of the following arithmetic series:
34 + 32 + 30 +...+10
Decide whether the following sequence is an A.P., if so find the 20th term of the progression:
–12, –5, 2, 9, 16, 23, 30, ..............
Given Arithmetic Progression 12, 16, 20, 24, . . . Find the 24th term of this progression.
Find the 19th term of the following A.P.:
7, 13, 19, 25, ...
Find the 27th term of the following A.P.
9, 4, –1, –6, –11,...
Select the correct alternative and write it.
What is the sum of first n natural numbers ?
Find the 23rd term of the following A.P.: 9, 4,-1,-6,-11.
Choose the correct alternative answer for the following sub-question
If the third term and fifth term of an A.P. are 13 and 25 respectively, find its 7th term
Find tn if a = 20 आणि d = 3
Find t5 if a = 3 आणि d = −3
Decide whether the given sequence 24, 17, 10, 3, ...... is an A.P.? If yes find its common term (tn)
Decide whether 301 is term of given sequence 5, 11, 17, 23, .....
Activity :- Here, d = `square`, therefore this sequence is an A.P.
a = 5, d = `square`
Let nth term of this A.P. be 301
tn = a + (n – 1) `square`
301 = 5 + (n – 1) × `square`
301 = 6n – 1
n = `302/6` = `square`
But n is not positive integer.
Therefore, 301 is `square` the term of sequence 5, 11, 17, 23, ......
12, 16, 20, 24, ...... Find 25th term of this A.P.
If tn = 2n – 5 is the nth term of an A.P., then find its first five terms
The nth term of an A.P. 5, 8, 11, 14, ...... is 68. Find n = ?
If p - 1, p + 3, 3p - 1 are in AP, then p is equal to ______.
