Question

if in an AP the sum of 12 terms is equal to 18 and the sum of 18 terms is equal to 12 then prove that the sum of 30 terms is minus 30 ......answer in pic if possible

Answer 1

The answer is explained in the attachment.


Hope it helps!

Answer 2

Concept:

Arithmetic Progression i.e. AP is a sequence of numbers in a specific order, in which the difference between any two consecutive numbers is a constant for the whole sequence.

Given:

The sum of an AP of 12 terms = 18,

And,

The sum of an AP of 18 terms = 12,

Find:

We are asked to prove that the sum of 30 terms is minus 30.

Solution:

We have,

The sum of an AP of 12 terms = 18,

And,

The sum of an AP of 18 terms = 12,

So,

Using the Sum of AP formula,

i.e.

Sₙ = (n/2)[2a + (n - 1)d]

So,

Here,

n = 12 and Sₙ = 18

So,

18 = (12/2)[2a + (12 - 1)d]

18 = 6 × [2a + 11d]

We get,

2a + 11d = 3     .....(i)

NOw,

n = 18 and Sₙ = 12

So,

12 = (18/2)[2a + (18 - 1)d]

12 = 9 × [2a + 17d]

We get,

2a + 17d = 4/3     .....(ii)

NOw,

Adding (i) and (ii),

2a + 11d = 3

2a + 17d = 4/3

We get,

-6d = 5/3

i.e.

d = -(5/18)

Now,

Putting d = -(5/18) in (ii),

i.e.

2a + 17 × (-5/18)  = 4/3

We get,

2a = 109/18

⇒

a = 109/36

NOw,

n = 30 and a = 109/36 and d = (-5/18)

i.e.

Sₙ = (n/2)[2a + (n - 1)d]

i.e.

Sₙ = (30/2)[2 × (109/36) + (30 - 1) × (-5/18)]

Sₙ = 15[ (109/18) + 29 × (-5/18)]

We get,

Sₙ = 15 × (-) = -30

So,

It is proven that the sum of 30 terms is minus 30.

Hence, it is proven that the sum of 30 terms is minus 30.

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