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An arithmetic progression has first term 2 and non-zero common difference, and the ratio S2n/Sn of the sum of its first 2n terms to that of its first…

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An arithmetic progression has first term

2

and non-zero common difference, and the ratio

\dfrac{S_{2n}}{S_n}

of the sum of its first

2n

terms to that of its first

n

terms is the same for every positive integer

n

. The

10^{\text{th}}

term of the progression is

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Answer: (A) $38$

For first term

2

and common difference

d

S_n=\dfrac n2\,(4+(n-1)d)

. With

k=4-d

this gives

\dfrac{S_{2n}}{S_n}=\dfrac{2\,(2nd+k)}{nd+k}=4-\dfrac{2k}{nd+k}

. Since

d\ne0

, the term

\dfrac{2k}{nd+k}

changes with

n

unless

k=0

. So

4-d=0

, i.e.

d=4

, and then

S_n=2n^2

with ratio

4

for every

n

. Therefore

a_{10}=2+9\cdot4=38

. As a check, the first few terms are

2,6,10,14,\ldots

and

S_1=2

S_2=8

S_4=32

, so

S_2/S_1=S_4/S_2=4

, consistent with the constant ratio.

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