If sum of the four consecutive terms of an A.P. is 32 and the ratio of the product of the first and
the last term to the product of the middle two terms is 7:15, find the terms.
Answers
Step-by-step explanation:
Let the four consecutive numbers in A.P. be a−3d,a−d,a+d,a+3d
where a is the first term and d is the common difference of the A.P.
According to the given question,
a−3d+a−d+a+d+a+3d=32
⇒4a=32
⇒a=8…(i)
and
(a−d)(a+d)
(a−3d)(a+3d)
=
15
7
⇒
(a
2
−d
2
)
(a
2
−(3d)
2
)
=
15
7
[∵(a+b)(a−b)=a
2
−b
2
]
⇒15(a
2
−9d
2
)=7(a
2
−d
2
)
⇒15a
2
−135d
2
=7a
2
−7d
2
⇒8a
2
=128d
2
⇒8(8)
2
=128d
2
[From(i)]
⇒d
2
=
128
8×8×8
⇒d
2
=4
⇒d=±2
Case 1: When d=2
The numbers are 8−(3×2),8−2,8+2,8+(3×2)
=2,6,10,14
Case 2: When d=−2
The numbers are 8−(3×−2),8−(−2),8+(−2),8+(3×−2)
=14,10,6,2