Question

There is a sequence of 4 positive integer terms, the first three are in AP. the last three are
in G. P. where the 4th term exceeds the first term by 30, then the sum of all the terms is

• A. $127$
• B. $128$
• C. $129$
• D. None of these

Answer 1

The correct option is C $129$

Let $a-d,a,a+d,a-d+30$ be the sequence of $4$ positive integer terms.
Given that the last three terms are in G.P.

That is, $a,a+d,a-d+30$ are in G.P.
$\Rightarrow {(a+d)}^2=a(a-d+30)$
$\Rightarrow a^2+2ad+d^2=a^2-ad+30a$
$\Rightarrow 3ad=30a-d^2$
$\Rightarrow 3a=\frac{d^2}{10-d}$
Therefore, R.H.S should be a multiple of $3$.
Now, For $d=3$, R.H.S is $\frac{9}{7}$; which is not a multiple of $3$.
Also, for $d=6$, R.H.S is $9$.
$\Rightarrow a=3$; but in this case $a-d$ is negative, which is not possible.
Now, for $d=9$, R.H.S is $81$.
$\Rightarrow a=27$
Since $d=12$ and afterwards make R.H.S as negative.
Therefore, $a=27$ and $d=9$.
Thus, the numbers are $18,27,36,48.$
Hence, the sum of $18,27,36,48$ is $129.$