Question
There are four numbers of which the first three are in GP and the last three are in AP, whose common difference is 6 . If the first and the last numbers are equal, then two other numbers are
BITSAT
Mathematics
Sequences and Series
2024
BITSAT 2024 (Memory Based Paper 3)
BITSAT Mathematics Question (2024) — Solution
Options
- A. -2,4
- B. -4,2
- C. 2,6
- D. None of the above
Answer
B. -4,2
Step-by-step solution
Let 3 numbers in AP are a,(a+6),(a+12) Also, first and last number out of 4 numbers are equal 4 numbers are (a+12), a,(a+6),(a+12) Also, given, first 3 numbers are in GP array ll & a^2=(a+12)(a+6) \\ & a^2=a^2+18 a+72 \\ & a=- 72 18 =-4 array 4 numbers are 8,-4,2,8
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