If a1, a2, a3…. and b1, b2, b3….. are A.P., and a1 = 2, a10 = 3, a1b1 = 1 – a10b10, then a4b4 is equal to……...

a 1 , a 2 , a 3 . . . . . . are in A.P. (Let common difference is d1) b 1 , b 2 , b 3 . . . . are in A.P. (Let common difference is d2)and a12, a10 = 3, a1b1 = 1 = a10b10 ? a 1 b 1 = 1  ∴ b 1 = 1 2 a 1 0 b 1 0 = 1  ∴ b 1 0 = 1 3   Now, a 1 0 = a 1 + 9 d 1 ⇒ d 1 = 1 9  b 1 0 = b 1 + 9 d 2 ⇒ d 2 = 1 9 [ 1 3 − 1 2 ] = − 1 5 4  Now a 4 = 2 + 3 9 = 7 3  b 4 = 1 2 − 3 5 4 = 4 9 ∴ a 4 b 4 = 2 8 2 7

If a1, a2, a3…. and b1, b2, b3….. are A.P., and a1 = 2, a10 = 3, a1b1 = 1 – a10b10, then a4b4 is equal to……...

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Maths Class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Nine Sequences and Series

If a₁, a₂, a₃…. and b₁, b₂, b₃….. are A.P., and a₁ = 2, a₁₀ = 3, a₁b₁ = 1 – a₁₀b₁₀, then a₄b₄ is equal to……...

Option 1 -

$\frac{35}{27}$

Option 2 -

Option 3 -

$\frac{27}{28}$

Option 4 -

$\frac{28}{27}$

3 Views | Posted 2 months ago

Asked by Shiksha User

1 Answer

Answered by

Payal Gupta | Contributor-Level 10

2 months ago

Correct Option - 4

Detailed Solution:

$a_{1}, a_{2}, a_{3} . . . . . .$ are in A.P. (Let common difference is d₁)

$b_{1}, b_{2}, b_{3} . . . .$ are in A.P. (Let common difference is d₂)

and a₁2, a₁₀ = 3, a₁b₁ = 1 = a₁₀b₁₀

$? a_{1} b_{1} = 1$

$∴ b_{1} = \frac{1}{2}$

$a_{10} b_{10} = 1$

$∴ b_{10} = \frac{1}{3}$

Now,

$a_{10} = a_{1} + 9 d_{1} \Rightarrow d_{1} = \frac{1}{9}$

$b_{10} = b_{1} + 9 d_{2} \Rightarrow d_{2} = \frac{1}{9} [\frac{1}{3} - \frac{1}{2}] = - \frac{1}{54}$

Now

$a_{4} = 2 + \frac{3}{9} = \frac{7}{3}$

$b_{4} = \frac{1}{2} - \frac{3}{54} = \frac{4}{9}$

$∴ a_{4} b_{4} = \frac{28}{27}$

...more

$a_{1}, a_{2}, a_{3} . . . . . .$ are in A.P. (Let common difference is d₁)

$b_{1}, b_{2}, b_{3} . . . .$ are in A.P. (Let common difference is d₂)

and a₁2, a₁₀ = 3, a₁b₁ = 1 = a₁₀b₁₀

$? a_{1} b_{1} = 1$

$∴ b_{1} = \frac{1}{2}$

$a_{10} b_{10} = 1$

$∴ b_{10} = \frac{1}{3}$

Now,

$a_{10} = a_{1} + 9 d_{1} \Rightarrow d_{1} = \frac{1}{9}$

$b_{10} = b_{1} + 9 d_{2} \Rightarrow d_{2} = \frac{1}{9} [\frac{1}{3} - \frac{1}{2}] = - \frac{1}{54}$

Now

$a_{4} = 2 + \frac{3}{9} = \frac{7}{3}$

$b_{4} = \frac{1}{2} - \frac{3}{54} = \frac{4}{9}$

$∴ a_{4} b_{4} = \frac{28}{27}$

less

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If a₁, a₂, a₃…. and b₁, b₂, b₃….. are A.P., and a₁ = 2, a₁₀ = 3, a₁b₁ = 1 – a₁₀b₁₀, then a₄b₄ is equal to……...

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