The next (4??) term of the AP

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Solution

Q.
The next (4th) term of the AP $\sqrt 18, \sqrt 50, \sqrt 98, \dots i s :$

1 $\sqrt 128$

2 $\sqrt 140$

3 $\sqrt 162$

4 $\sqrt 200$

Ans.
(3)

$W e h a v e, \sqrt 18 = \sqrt 9 \times 2 = 3 \sqrt 2 \sqrt 50 = \sqrt 25 \times 2 = 5 \sqrt 2 a n d \sqrt 98 = \sqrt 49 \times 2 = 7 \sqrt 2 \Rightarrow 3 \sqrt 2, 5 \sqrt 2, 7 \sqrt 2, \dots a r e i n A P w h e r e a = 3 \sqrt 2 a n d d = 7 \sqrt 2 - 5 \sqrt 2 = 2 \sqrt 2 T h u s, 4 ᵗ ʰ t e r m = 3 \sqrt 2 + (4 - 1) \times 2 \sqrt 2 = 3 \sqrt 2 + 6 \sqrt 2 = 9 \sqrt 2 = \sqrt (9) ² \times 2 = \sqrt 162 O p t i o n (c) i s c o r r e c t .$

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