the sum of the first n terms of an AP is half the sum of the next n terms of - Maths - Arithmetic Progressions

Question

the sum of the first n terms of an AP is half the sum of the next n
terms of the same progression, then prove that the ratio between
the sum of the first 3n terms and the sum of first n terms is 6

Varun.Rawat · Accepted Answer

According to the question,
Sum of 1st n terms = 1/2 of sum of 2nd n terms
$\frac{n}{2} (2 a + (n - 1) d) = \frac{1}{2} (\frac{n}{2} (2 a_{n + 1} + (n - 1) d)) 4 a + 2 (n - 1) d = 2 a_{n + 1} + (n - 1) d 4 a + (n - 1) d = 2 (a + (n + 1 - 1) d) 4 a + (n - 1) d = 2 a + 2 n d \Rightarrow 2 a = (n + 1) d$
Now, Required ratio
$= \frac{S_{3 n}}{S_{n}} = \frac{\frac{3 n}{2} (2 a + (3 n - 1) d)}{\frac{n}{2} (2 a + (n - 1) d)} = \frac{3 ((n + 1) d + (3 n - 1) d)}{(n + 1) d + (n - 1) d} = \frac{3 d (n + 1 + 3 n - 1)}{d (n + 1 + n - 1)} = \frac{3 (4 n)}{2 n} = \frac{6}{1}$

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the sum of the first n terms of an AP is half the sum of the next n terms of the same progression, then prove that the ratio between the sum of the first 3n terms and the sum of first n terms is 6.