Question

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 ?

Answer 1

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 2

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