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The sum of the first 7 terms of an arithmetic progression (AP) is 182. If the fourth term and the 17th term are in the ratio 1:5, find the AP.

Answer :

The AP is 34, 30, 26, ..., whose first term is 34 and common difference is -4.

Let the first term of the AP be a and the common difference be d.

We are given that the sum of the first 7 terms of the AP is 182.

The sum of the first n terms of an AP is given by the formula:

S\_n = n/2[2a + (n-1)d]

For n = 7, we have:

182 = 7/2[2a + 6d]

Simplifying, we get:

a + 3d = 26 ...(1)

We are also given that the fourth term and the 17th term of the AP are in the ratio 1:5.

The fourth term of the AP is a + 3d and the 17th term is a + 16d.

So, we have:

(a + 3d) / (a + 16d) = 1/5

Simplifying, we get:

4a + 13d = 0 ...(2)

Now, we have two equations (1) and (2) with two variables a and d.

We can solve these equations simultaneously to find the values of a and d.

From (1), we get:

a = 26 - 3d

Substituting this value in (2), we get:

4(26 - 3d) + 13d = 0

Simplifying, we get:

d = -4

Substituting the value of d in (1), we get:

a = 26 - 3(-4) = 34

Therefore, the AP is given by:

a, a+d, a+2d, ...

That is, 34, 34-4, 34-8, ...

The correct answer is option C: The AP is 34, 30, 26, ...

Numeric calculations:

From (1), we get:

a = 26 - 3d

Substituting this value in (2), we get:

4(26 - 3d) + 13d = 0

Simplifying, we get:

104 - 12d + 13d = 0

d = -4

Substituting the value of d in (1), we get:

a = 26 - 3(-4) = 34

Therefore, the AP is given by:

a, a+d, a+2d, ...

That is, 34, 34-4, 34-8, ...

So, the AP is 34, 30, 26, ...

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