Question

Parikh is teaching an interesting question. It is about decreasing arithmetic progression. He says- “Let us take any decreasing arithmetic progression where sum of all its terms, except for the first term, is equal to -36. And sum of all its terms, except for the last term, is zero. And the difference of the tenth and the sixth term is equal to -16, then what will be the first term of this series.” Can you solve this question.
16

20

-16

-20

-24


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Answer 1

Let us consider the decreasing arithmetic progression to be of n terms:

a, a - r, a - 2r, a - 3r, ..., a - r (n - 2), a - r (n - 1).

Condition 1. Sum of all its items except for the first term, is equal to (- 36)

Then the series contains (n - 1) terms.

So its sum be

= (n - 1)/2 * [a - r + a - r (n - 1)]

= (n - 1)/2 * [a - r + a - rn + r]

= (n - 1)/2 * [2a - rn]

Given, (n - 1)/2 * [2a - rn] = - 36 .....(1)

Condition 2. Sum of all its items except for the last term, is equal to 0.

Then the series contains (n - 1) terms.

So its sum be

= (n - 1)/2 * [a + a - r (n - 2)]

= (n - 1)/2 * [a + a - rn + 2r]

= (n - 1)/2 * [2a + 2r - rn]

Given, (n - 1)/2 * [2a + 2r - rn] = 0

or, 2a + 2r - rn = 0 .....(2)

Condition 3. The difference of the tenth and the sixth term is equal to (- 16).

Here, tenth term = a - 9r

and sixth term = a - 5r

Given, (a - 9r) - (a - 5r) = - 16

or, a - 9r - a + 5r = - 16

or, - 4r = - 16

or, r = 4

Putting r = 4 in (2), we get

2a + 2 (4) - (4) n = 0

or, 2a + 8 - 4n = 0

or, 2a - 4n = - 8 .....(3)

Putting r = 4 in (1), we get

(n - 1)/2 * [2a - 4n] = - 36

or, (n - 1)/2 * (- 8) = - 36, using (3)

or, - 4 (n - 1) = - 36

or, n - 1 = 9

or, n = 10

We put r = 4 and n = 10 in (2):

2a + 2 (4) - 4 (10) = 0

or, 2a + 8 - 40 = 0

or, 2a = 32

or, a = 16

∴ the first term of the arithmetic progression is 16.

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