Question
Let a_1, a_2, a_ 40 be in AP and h_1, h_2, . h_ 10 be in HP. If a_1=h_1=2 and a_ 10 =h_ 10 =3, then a_4 h_7 is
BITSAT
Mathematics
Sequences and Series
2022
BITSAT 2022
BITSAT Mathematics Question (2022) — Solution
Options
- A. 2
- B. 3
- C. 5
- D. 6
Answer
D. 6
Step-by-step solution
Let d be the common difference of the AP. Then, aligned & a_ 10 =3 a_1+9 d=3 \\ & 2+9 d=3 d= 1 9 \\ & a_4=a_1+3 d=2+ 1 3 = 7 3 aligned Let D be the common difference of 1 h_1 , 1 h_2 , .- 1 h_ 10 . Then, h_ 10 =3 array ll & 1 h_ 10 = 1 3 1 2 +9 D= 1 3 \\ & 9 D=- 1 6 D=- 1 54 \\ & 1 h_7 = 1 h_1 +6 D= 1 2 - 1 9 = 7 18 \\ & h_7= 18 7 \\ & a_4 h_7= 7 3 18 7 =6 array
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