IPMAT Indore 2026 · Quantitative Ability · Question 8

Step 1 — write the condition. With first term a and common difference d:

a+\left( n-1 \right)d=174 \Rightarrow \left( n-1 \right)d=174-a

Since 174>a in both cases, d must be positive; the even positive values below 10 are d\in \{2,4,6,8\}.

Step 2 — case a=132. Then \left( n-1 \right)d=42:

d	42/d	n
2	21	22
4	10.5	not an integer
6	7	8
8	5.25	not an integer

Step 3 — case a=158. Then \left( n-1 \right)d=16:

d	16/d	n
2	8	9
4	4	5
6	2.67	not an integer
8	2	3

Step 4 — collect the distinct values.

n\in \{22, 8, 9, 5, 3\}\Rightarrow 5\text{values}

Answer: 5

Shortcut. This is a divisor hunt, not an AP question: you need d to divide 174-a. So compute 42 and 16, then just ask which of 2,4,6,8 divide each. Two divisors for 42, three for 16.