SATHEE: Sequences and Series 5 Question 12

Sequences and Series 5 Question 12

3. Let $a_{1}, a_{2}, \dots, a_{10}$ be in $A P$ and $h_{1}, h_{2}$ , equal to $\dots . ., h_{10}$ be in HP. If $a_{1} = h_{1} = 2$ and $a_{10} = h_{10} = 3$ , then $a_{4} h_{7}$ is

$(1999, 2 M)$

(a) 2

(b) 3

(d) 6

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Answer:

Correct Answer: 3. (d)

Solution:

Since, $a_{1}, a_{2}, a_{3}, \dots, a_{10}$ are in AP.

Now, $a_{10} = a_{1} + 9 d$

$\Rightarrow 3 = 2 + 9 d$

$\Rightarrow d = 1 / 9$ and $a_{4} = a_{1} + 3 d$

$\Rightarrow a_{4} = 2 + 3 (1 / 9) = 2 + 1 / 3 = 7 / 3$

Also, $h_{1}, h_{2}, h_{3}, \dots, h_{10}$ are in HP.

$\Rightarrow \frac{1}{h_{1}}, \frac{1}{h_{2}}, \frac{1}{h_{3}}, \dots, \frac{1}{h_{10}}$ are in AP.

Given, $h_{1} = 2, h_{10} = 3$

$∴ \frac{1}{h_{10}} = \frac{1}{h_{1}} + 9 d_{1} \Rightarrow \frac{1}{3} = \frac{1}{2} + 9 d_{1}$

$\Rightarrow - \frac{1}{6} = 9 d_{1}$

$\Rightarrow d_{1} = - \frac{1}{54}$ and $\frac{1}{h_{7}} = \frac{1}{h_{1}} + 6 d_{1}$

$\begin{array}{ll} \Rightarrow & \frac{1}{h_{7}} = \frac{1}{2} + \frac{6 \times 1}{- 54} \\ \Rightarrow & \frac{1}{h_{7}} = \frac{1}{2} - \frac{1}{9} \Rightarrow h_{7} = \frac{18}{7} \\ ∴ & a_{4} h_{7} = \frac{7}{3} \times \frac{18}{7} = 6 \end{array}$

Sequences and Series 5 Question 12

3. Let $a_{1}, a_{2}, \dots, a_{10}$ be in $A P$ and $h_{1}, h_{2}$ , equal to $\dots . ., h_{10}$ be in HP. If $a_{1} = h_{1} = 2$ and $a_{10} = h_{10} = 3$ , then $a_{4} h_{7}$ is

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Sequences and Series 5 Question 12

3. Let a1,a2,…,a10 be in AP and h1,h2, equal to …..,h10 be in HP. If a1=h1=2 and a10=h10=3, then a4h7 is

Answer:

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Sample Mock Test

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3. Let $a_{1}, a_{2}, \dots, a_{10}$ be in $A P$ and $h_{1}, h_{2}$ , equal to $\dots . ., h_{10}$ be in HP. If $a_{1} = h_{1} = 2$ and $a_{10} = h_{10} = 3$ , then $a_{4} h_{7}$ is