CAT 2025 Quant was dominated by Algebra followed by Arithmetic. In Arithmetic, the questions were dominated by topics like Speed-time-distance, Mixture and Alligations. This year, there was a surprise. The questions from Geometry were relatively on the lower side as compared to the previous years. There were 8 TITA Qs this year. Overall this section was at a medium level of difficulty.
Question 8 : Let \( a _ { n } \) be the \( n ^ { \text {th } } \) term of a decreasing infinite geometric progression. If \( a _ { 1 } + a _ { 2 } + a _ { 3 } = 52 \) and \( a _ { 1 } a _ { 2 } + a _ { 2 } a _ { 3 } + a _ { 3 } a _ { 1 } = 624 \), then the sum of this geometric progression is
Let terms be \(a,ar,ar^2\):
\(a+ar+ar^2=52\)
\(a^2r+a^2r^2+a^2r^3=624\)
\(\Rightarrow ar(a+ar+ar^2)=624\)
Solve for second term \(a_2=ar\)
\(ar\times52=624\Rightarrow ar=12\)
Solve for first term \(a\) and common ratio \(r\)
Since \(ar=12\Rightarrow a=\frac{12}{r}\):
\(\frac{12}{r}+12+12r=52\)
\(12+12r^2-40r=0\)
\(3r^2-10r+3=0\)
\((3r-1)(r-3)=0\)
Since it is a decreasing GP \((|r|<1)\), \(r=\frac{1}{3}\).
\(a=\frac{12}{1/3}=36\)
Compute infinite sum
\(S=\frac{a}{1-r}\)
\(=\frac{36}{1-\frac{1}{3}}\)
\(=\frac{36}{\frac{2}{3}}=54\)
Choice 3 is the correct answer.
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