Let a1-a2-a3- a21 be an AP such that -n-1201anan-1-49 If the sum of this AP is 189- then a6a16 is equal to-

Let first term be a and common difference d 1a_1a_2+1a_2a_3+⋯+1a_20a_21=49⋯(i) Also, a_1+a_2+⋯+a_21=189⋯(ii) by (i) 1a_1 1a_2+1a_2 1a_3+⋯+1a_20 1a_21=4d9 ⇒1a 1a ...

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Byju's Answer

Standard XII

Mathematics

Geometric Progression

Let a1,a2,a3⋯...

Question

Let $a_{1}, a_{2}, a_{3} \dots, a_{21}$ be an $A P$ such that $20 \sum n = 1 \frac{1}{a_{n} a_{n + 1}} = \frac{4}{9}$ If the sum of this $A P$ is $189$ , then $a_{6} a_{16}$ is equal to:

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72

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Solution

The correct option is C $72$
Let first term be $a$ and common difference $d$
$\frac{1}{a_{1} a_{2}} + \frac{1}{a_{2} a_{3}} + \dots + \frac{1}{a_{20} a_{21}} = \frac{4}{9} \dots (i)$
Also, $a_{1} + a_{2} + \dots + a_{21} = 189 \dots (i i)$
by $(i)$
$\frac{1}{a_{1}} - \frac{1}{a_{2}} + \frac{1}{a_{2}} - \frac{1}{a_{3}} + \dots + \frac{1}{a_{20}} - \frac{1}{a_{21}} = \frac{4 d}{9}$
$\Rightarrow \frac{1}{a} - \frac{1}{a + 20 d} = \frac{4 d}{9}$
$\Rightarrow \frac{20 d}{a (a + 20 d)} = \frac{4 d}{9} \Rightarrow 45 = a (a + 20 d) \dots (i i i)$
From $(i i)$
$21 a + 210 d = 189 \Rightarrow a + 10 d = 9 \dots (i v)$
by $(i i i)$ and $(i v)$
$d = \frac{3}{5} and a = 3$
$∴ a_{6} a_{16} = (3 + 3) (3 + 9) = 72$

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Let a1,a2,a3⋯,a21 be an AP such that 20∑n=11anan+1=49 If the sum of this AP is 189, then a6a16 is equal to:

Let $a_{1}, a_{2}, a_{3} \dots, a_{21}$ be an $A P$ such that $20 \sum n = 1 \frac{1}{a_{n} a_{n + 1}} = \frac{4}{9}$ If the sum of this $A P$ is $189$ , then $a_{6} a_{16}$ is equal to: