IPMAT Indore 2022 · Quantitative Ability · Question 24

Step 1 — use the block property of an AP. Pair each term of the second block with the corresponding term of the first: term 16 with term 1, term 17 with term 2, and so on. Each pair differs by exactly 15d, because

{T}_{16}-{T}_{1}=15d,{T}_{17}-{T}_{2}=15d, \ldots

Step 2 — sum those 15 differences.

\left( \text{sum of next}15 \right)-\left( \text{sum of first}15 \right)=15\times 15d=225d

Step 3 — substitute the given sums.

350-200=150=225d \Rightarrow d=\frac{150}{225}=\frac{2}{3}

Answer: Option 2 — \frac{2}{3}

Shortcut. Memorise the block rule: for an AP, consecutive blocks of k terms differ by {k}^{2}d.

Here k=15, so the gap between the two blocks is 225d, and

d=\frac{350-200}{225}=\frac{150}{225}=\frac{2}{3}

This avoids the standard route entirely — no need to set up 15/2\left( 2a+14d \right)=200 and a second equation for 30 terms, then eliminate a. The block rule is one subtraction and one division, and it generalises: three consecutive blocks of k terms are themselves in AP with common difference {k}^{2}d.