# The sum of the fi

Given sum of first seven terms of AP, S7 = 182.

We know that sum of first n terms of AP, Sn = (n/2) [2a + (n – 1) d]

182 = (7/2) [2a + (7 – 1) d]

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

182 = (7/2) (2) [a + 3d]

182/7 = a + 3d

a + 3d = 26 … (1)

Also given the 4th and 17th terms are in the ratio 1: 5.

We know that nth term, an = a + (n – 1) d

a4 = a + (4 – 1) d = a + 3d

a17 = a + (17 – 1) d = a + 16d

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

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

5a + 15d = a + 16d

4a = d … (2)

Substituting the above in (1),

a + 3 (4a) = 26

a + 12a = 26

13a = 26

a = 2

Substituting a = 2 in (2),

d = 8

The AP will be as follows: 2, 10, 18, 26, …

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