The fifth term of the H-P -2- 212- 313- will be

2,52,103,.... is a H.P ⇒12,25,310,....is an A.P. Fifth term of A.P=t_5=a+4d a=12,d=25 12=4 510= 110 ⇒ a_5=a+4d=12+4( 110) ⇒ a_5=12 410 nbsp; nbsp; n ...

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Standard X

Mathematics

Putting It Together

The fifth ter...

Question

The fifth term of the H.P :
$2, 2 \frac{1}{2}, 3 \frac{1}{3}, . . . . . .,$ will be

5 \frac{1}{5}

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3 \frac{1}{5}

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\frac{1}{10}

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10

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Solution

The correct option is D
$10$
$2, \frac{5}{2}, \frac{10}{3}, . . . . i s a H . P \Rightarrow \frac{1}{2}, \frac{2}{5}, \frac{3}{10}, . . . . i s a n A . P . F i f t h t e r m o f A . P = t_{5} = a + 4 d a = \frac{1}{2}, d = \frac{2}{5} - \frac{1}{2} = \frac{4 - 5}{10} = \frac{- 1}{10} \Rightarrow a_{5} = a + 4 d = \frac{1}{2} + 4 (\frac{- 1}{10}) \Rightarrow a_{5} = \frac{1}{2} - \frac{4}{10} = \frac{5 - 4}{10} = \frac{1}{10} T h u s, f i f t h t e m o f H . P = \frac{1}{t_{5}} = \frac{1}{1 / 10} = 10$
So, option d is correct.

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The fifth term of the H.P : 2, 212, 313,......, will be

The correct option is D 102,52,103,.... is a H.P⇒12,25,310,....is an A.P.Fifth term of A.P=t5=a+4da=12,d=25−12=4−510=−110⇒a5=a+4d=12+4(−110)⇒a5=12−410 =5−410=110Thus, fifth tem of H.P=1t5=11/10=10 So, option d is correct.

The fifth term of the H.P :
$2, 2 \frac{1}{2}, 3 \frac{1}{3}, . . . . . .,$ will be