If the sum of first 11 terms of an arithmetic progression equals that of the first 19 terms,then what is the?

Question

If the sum of first 11 terms of an arithmetic progression equals that of the first 19 terms,then what is the sum of first 30 terms.
(a) 0
(b) -1
(c) 1
(d) Not unique

Answer 1

Let s,n denote the sum of the 1st n terms of an arithmetic progression.
Let a,n denote the nth term of the arithmetic progression
Let n = number of terms of the arithmetic progression
Let d = the common difference of the arithmetic progression

We have the formulae
a,n = a,1 + (n - 1)d ........ (1)
s,n = n(a,1 + a,n)/2 ......... (2)


SOLUTION

It says that
s,11 = s,19

Use (2) for s,11 and s,19. Thus,
11(a,1 + a,11)/2 = 19(a,1 + a,19)/2

Multiply 2 to both sides
11(a,1 + a,11) = 19(a,1 + a,19)

Use (1) for a,11 and a,19. Thus,
11[a,1 + a,1 + (11 - 1)d]= 19(a,1 + a,1 + (19 - 1)d]

Combine like terms
11(2a,1 + 10d) = 19(2a,1 + 18d)

Divide both sides by 2 (bec. all terms inside the parentheses are divisible by 2)
11(a,1 + 5d) = 19(a,1 + 9d)

Distribute
11a,1 + 55d = 19a,1 + 171d

Transpose everything to the right
8a,1 + 116d = 0

Divide by 4
2a,1 + 29d = 0 ----- let it be identity (3)


We are asked to find s,30
s,30 = 30(a,1 + a,30)/2

Simplify and subs. a,30
s,30 = 15[a,1 + a,1 + (30 - 1)d]

Add
s,30 = 15(2a,1 + 29d)

Subs. the expression in the parentheses with identity (3)
s,30 = 15(0)

Therefore,
s,30 = 0

The sum is therefore 0. And the correct answer is choice (a).


NOTES:
Although the answer is 0, this doesn't mean that the whole progression are zeroes. A good example is the arithmetic progression

29 27 25 23 21 19 ...

Here the sum of the 1st 11 terms and the 1st 19 terms are both 209, but still the sum of the 1st 30 terms is zero.

^_^

^_^

Answer 2

(a) 0

sum to 11 of AP = sum to 19 of AP
means sum of 12 to 19 terms = 0.
For an AP to achieve this, it passes 0 from 15th to 16th term, and the nth term = negative of the (30-n+1)th term.
So, sum to 30 = (1st term + 30th term) + (2nd+29th) + ... +(15th + 16th) = 0.

Answer 3

the sum of n first terms of arithmetic progression a1,a2,a3,... is :

Sn=n*(2*a1+(n-1)*d)/2

S11=S19
=> 11*(2*a1+10d)/2 = 19*(2*a1+18d)/2

=> 22*a1 + 110*d= 38*a1 + 342d

=> -16*a1 = 232*d

=> a1 = -14.5*d

The sum of the first 30 terms is :

S30=30*(2*a1+29d)/2 =30*a1+435*d

if we substitute a1 for -14.5*d, we get:

S30= 30*(-14.5*d)+435*d
= -435*d +435*d =0

so the answer is zero, but this does not mean that the whole progression is zeroes.

Answer 4

Answer is 0.
Why? Simple logic.