IPMAT Indore 2022 · Quantitative Ability · Question 13

Step 1 — apply the AP condition. Three numbers are in AP when twice the middle equals the sum of the outer two:

2\left({2}^{x+3}-{2}^{2x-1}-16 \right)=-16+\left({2}^{2x-1}+16 \right)={2}^{2x-1}

Step 2 — expand the left side.

{2}^{x+4}-{2}^{2x}-32={2}^{2x-1}

Step 3 — substitute t={2}^{x} (so {2}^{x+4}=16t, {2}^{2x}={t}^{2}, {2}^{2x-1}={t}^{2}/2):

16t-{t}^{2}-32=\frac{{t}^{2}}{2}

Step 4 — clear the fraction and solve. Multiply by 2:

32t-2{t}^{2}-64={t}^{2}\Rightarrow 3{t}^{2}-32t+64=0

t=\frac{32\pm \sqrt{1024-768}}{6}=\frac{32\pm 16}{6}\Rightarrow t=8 \text{or}t=\frac{8}{3}

Step 5 — return to x. We need {2}^{x}=t with x a real number; the question expects an integer answer, and only

{2}^{x}=8 \Rightarrow x=3

gives one. (t=8/3 gives x={\log}_{2}8/3, not a clean value.)

Check at x=3: the three terms are -16, {2}^{6}-{2}^{5}-16=64-32-16=16, and {2}^{5}+16=48. Differences: 16-\left( -16 \right)=32 and 48-16=32 ✔

Answer: 3

Shortcut. Substitute t={2}^{x} before doing anything else. Every term is a power of 2, and the substitution turns the condition into the plain quadratic

3{t}^{2}-32t+64=0 \Rightarrow \left( t-8 \right)\left( 3t-8 \right)=0

Then keep only the root that is a power of 2, giving x=3.

Watch the exponent arithmetic: {2}^{2x-1}=1/2{\left({2}^{x}\right)}^{2}, not {\left({2}^{x}\right)}^{2}-1. Mishandling that half is the commonest slip in this question.