if the sum of the first 11 terms of an arithmetic progression equals that of the first 19 terms , then what is the sum of first 30 terms ?
Answers
Answer:
0
Step-by-step explanation:
Sum of 11 terms of an AP equals the sum of 19 terms of the same AP.
S11 = (11/2)[2a+10d] …(1)
S19 = (19/2)[2a+18d] …(2)
Equate (1) and (2)
(11/2)[2a+10d] = (19/2)[2a+18d], or
22a+110d = 38a+342d, or
16a = -232d, or a = -(232/16)d
S30 = (30/2)[2a+29d]
= 15[-(464d/16) + 29d]
= (15/3)[-29d+29d]
= 0
The sum of the 30 terms of the AP is 0.
Hope it helps u.
Answer:
0
Step-by-step explanation:
Let the first term be a and com. diff. between be d. Using S = (n/2)[2a + (n - 1)d]
=> sum of 11 terms = sum of 19 terms
=> (11/2) (2a + 10d) = (19/2) (2a + 18d)
=> 11a + 55d = 19a + 171d
=> 55d - 171d = 19a - 11a
=> - 116d = 8a
=> - 29d = 2a ...(1)
Hence, sum of 30 terms is:
= (30/2) (2a + 29d)
= (15)(-29d + 29d) {2d = - 29d, from (1)}
= 15(0)
= 0
Sum of first 30 terms is 0