Question:hard

Find the sum of the first 10 terms of the series \(2, -8, 32, -128, 512, \ldots\)

Show Hint

Notice the series is geometric with ratio $-4$; pairing consecutive terms turns it into a new, smaller geometric series.
Updated On: Jul 8, 2026
  • \(-419430\)
  • \(-419340\)
  • \(-414930\)
  • \(-413940\)
Show Solution

The Correct Option is A

Solution and Explanation

Step 1: The series is geometric with first term $a=2$ and common ratio $r=-4$: the 10 terms are $2,-8,32,-128,512,-2048,8192,-32768,131072,-524288$.
Step 2: Group the terms into 5 consecutive pairs: $P_1=2-8=-6$, $P_2=32-128=-96$, $P_3=512-2048=-1536$, $P_4=8192-32768=-24576$, $P_5=131072-524288=-393216$.
Step 3: Since $r^2=16$, each pair is exactly 16 times the previous pair (check: $-96/-6=16$). So $P_1,P_2,\ldots,P_5$ themselves form a GP with first term $-6$ and ratio $16$.
Step 4: Sum of these 5 pairs $=\dfrac{-6(16^5-1)}{16-1}=\dfrac{-6(1048576-1)}{15}=\dfrac{-6(1048575)}{15}=-6(69905)=-419430$.
\[\boxed{-419430}\]
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