Question

Let \(a_1, a_2, a_3, a_4\) and \(a_5\) be in A.P. with the common difference greater than 2. If \(\sum_{i=1}^{5}a_i=30\) and \(a_1\cdot a_2\cdot a_3\cdot a_4\cdot a_5=3840\), then \(a_1\cdot a_5\) is equal to :

• A. 82
• B. \(-128\)
• C. 128
• D. \(-82\)

Hint:

Take the five terms as \(a-2d,\ a-d,\ a,\ a+d,\ a+2d\); the sum gives the middle term, and the product gives \(d^2\).

Answer 1

The Correct Option is B

Let the terms be \(a-2d,\ a-d,\ a,\ a+d,\ a+2d\) with \(d>2\).
Sum \(=5a=30\), so \(a=6\).
Product: \(6(36-d^2)(36-4d^2)=3840\), so \((36-d^2)(36-4d^2)=640\).
Let \(u=d^2\): \(4u^2-180u+1296=640\), i.e. \(u^2-45u+164=0\), so \(u=\dfrac{45\pm37}{2}=41\) or \(4\).
\(d>2\) rejects \(d^2=4\), so \(d^2=41\).
\(a_1\cdot a_5=(6-2d)(6+2d)=36-4d^2=36-164=-128\).
Answer: (2) \(-128\)