Question

Let \(a_1,a_2,\ldots\ldots\ldots,a_{21}\) be an AP such that \(\sum_{n=1}^{20}\frac{1}{a_na_{n+1}}=\frac49\). If the sum of this AP is 189, then \(a_6a_{16}\) is equal to :

• A. \(57\)
• B. \(72\)
• C. \(48\)
• D. \(36\)

Hint:

The sum telescopes to \(\frac{20}{a_1a_{21}}\); combine with the sum of the AP to find \(a_1,a_{21}\).

Answer 1

The Correct Option is B

For an AP with common difference \(d\): \(\frac{1}{a_na_{n+1}}=\frac1d\left(\frac1{a_n}-\frac1{a_{n+1}}\right)\), so the sum \(=\frac1d\left(\frac1{a_1}-\frac1{a_{21}}\right)=\frac{a_{21}-a_1}{d\,a_1a_{21}}=\frac{20}{a_1a_{21}}=\frac49\Rightarrow a_1a_{21}=45\).
Sum of 21 terms \(=\frac{21}{2}(a_1+a_{21})=189\Rightarrow a_1+a_{21}=18\).
So \(\{a_1,a_{21}\}=\{3,15\}\) and \(d=\pm\frac{12}{20}=\pm\frac35\).
If \(a_1=3\), \(d=\frac35\): \(a_6=6\), \(a_{16}=12\), product 72. If \(a_1=15\), \(d=-\frac35\): \(a_6=12\), \(a_{16}=6\), product 72.
Answer: (2) 72