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Sequences and Series question

2022 · 27 Jun · Shift 2 · Q24
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Sequences and Series question

2022 · 27 Jun · Shift 2 · Q24

JEE MainMathematicsSequences and SeriesMCQ+4 / −1
If a1, a2, a3 ...... and b1, b2, b3 ....... are A.P., and a1 = 2, a10 = 3, a1b1 = 1 = a10b10, then a4 b4 is equal to -
  1. A
    3527{{35} \over {27}}2735​
  2. B
    1
  3. C
    2728{{27} \over {28}}2827​
  4. D
    2827{{28} \over {27}}2728​
View written solutionFree

Correct answer: D

  1. Find the A.P. ana_nan​

Given a1=2a_1=2a1​=2 and a10=3a_{10}=3a10​=3.

For an A.P., an=a1+(n−1)daa_n=a_1+(n-1)d_aan​=a1​+(n−1)da​ So, a10=2+9da=3a_{10}=2+9d_a=3a10​=2+9da​=3 9da=1  ⟹  da=199d_a=1 \implies d_a=\frac{1}{9}9da​=1⟹da​=91​ Hence, an=2+n−19a_n=2+\frac{n-1}{9}an​=2+9n−1​

In particular, a4=2+3⋅19=2+13=73a_4=2+3\cdot \frac{1}{9}=2+\frac{1}{3}=\frac{7}{3}a4​=2+3⋅91​=2+31​=37​

  1. Use the product conditions to find b1b_1b1​ and b10b_{10}b10​

Given: a1b1=1,a10b10=1a_1b_1=1, \quad a_{10}b_{10}=1a1​b1​=1,a10​b10​=1 Since a1=2a_1=2a1​=2 and a10=3a_{10}=3a10​=3, 2b1=1  ⟹  b1=122b_1=1 \implies b_1=\frac{1}{2}2b1​=1⟹b1​=21​ 3b10=1  ⟹  b10=133b_{10}=1 \implies b_{10}=\frac{1}{3}3b10​=1⟹b10​=31​

Now bnb_nbn​ is also an A.P. with first term 12\frac1221​ and 10th term 13\frac1331​.

  1. Find the A.P. bnb_nbn​

Let common difference of bnb_nbn​ be dbd_bdb​. Then, b10=b1+9dbb_{10}=b_1+9d_bb10​=b1​+9db​ 13=12+9db\frac13=\frac12+9d_b31​=21​+9db​ 9db=13−12=−169d_b=\frac13-\frac12=-\frac169db​=31​−21​=−61​ db=−154d_b=-\frac{1}{54}db​=−541​

Therefore, b4=b1+3db=12−354=12−118=818=49b_4=b_1+3d_b=\frac12-\frac{3}{54}=\frac12-\frac{1}{18}=\frac{8}{18}=\frac49b4​=b1​+3db​=21​−543​=21​−181​=188​=94​

  1. Compute a4b4a_4b_4a4​b4​

a4b4=73⋅49=2827a_4b_4=\frac73\cdot \frac49=\frac{28}{27}a4​b4​=37​⋅94​=2728​

  1. Match with the options

2827\frac{28}{27}2728​ This is Option D.

  1. Compare with stored correct answer

Stored correct answer: D

Our derived answer: D

So, the answer agrees with the stored correct answer.

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