IPMAT Indore 2022 · Quantitative Ability · Question 11

Step 1 — name the AP. Let the first term be a and the common difference d. The three terms are

{T}_{3}=a+2d,{T}_{14}=a+13d,{T}_{69}=a+68d

Step 2 — impose the GP condition. Three terms are in GP when the middle one squared equals the product of the outer two:

{\left( a+13d \right)}^{2}=\left( a+2d \right)\left( a+68d \right)

Step 3 — expand and simplify.

{a}^{2}+26ad+169{d}^{2}={a}^{2}+70ad+136{d}^{2}

33{d}^{2}=44ad \Rightarrow 3d=4a \left( d\ne 0,\text{since the terms are distinct}\right)

So a=\frac{3d}{4}.

Step 4 — write the three GP terms.

{T}_{3}=\frac{3d}{4}+2d=\frac{11d}{4},{T}_{14}=\frac{3d}{4}+13d=\frac{55d}{4},{T}_{69}=\frac{3d}{4}+68d=\frac{275d}{4}

The common ratio is \frac{55}{11}=5 ✔ (and \frac{275}{55}=5 ✔).

Step 5 — find the next GP term and locate it in the AP.

\text{next}=\frac{275d}{4}\times 5=\frac{1375d}{4}

Setting this equal to a+\left( n-1 \right)d:

\frac{3d}{4}+\left( n-1 \right)d=\frac{1375d}{4}\Rightarrow \frac{3}{4}+n-1=\frac{1375}{4}

n=\frac{1375}{4}+\frac{1}{4}=\frac{1376}{4}=344

Answer: 344

Shortcut. Scale d=4 and the fractions vanish. Then a=3 and the AP is 3,7,11,\ldots with the three terms

{T}_{3}=11,{T}_{14}=55,{T}_{69}=275

— visibly a GP of ratio 5. The next GP term is 275\times 5=1375, and

3+\left( n-1 \right)4=1375 \Rightarrow n-1=343 \Rightarrow n=344

Choosing a convenient d is legitimate because the answer n is a position, not a value, so it cannot depend on the scale. Note the neat 343={7}^{3} at the end — a sign you have it right.