Progression (Sequence and Series)Hard
Question
Let a1,a2 , ......a10 in AP and h1, h2, ...... h10 be in HP. If a1 = h1 = 2 and a10 = h10 = 3, then a4h7 is
Options
A.2
B.3
C.5
D.6
Solution
Since, a1, a2,..... a10 be in AP
Now, a10 = a1 + 9d
⇒ 3 = 2 + 9d
⇒ d = 1 / 9
and a4 = aa1 + 3d
⇒ a4 = 2 + 3(1/ 9) = 2 +1/ 3 = 7 / 3
Also, h1, h2, h3 .....h10 in HP.
⇒
be in AP.
Given, h1 = 2, h10 = 3
∴
⇒ -
= 9d1 ⇒ d1 = - 
and
⇒
⇒ 
⇒
∴ a4h7 =
= 6
Now, a10 = a1 + 9d
⇒ 3 = 2 + 9d
⇒ d = 1 / 9
and a4 = aa1 + 3d
⇒ a4 = 2 + 3(1/ 9) = 2 +1/ 3 = 7 / 3
Also, h1, h2, h3 .....h10 in HP.
⇒
be in AP.Given, h1 = 2, h10 = 3
∴
⇒ -
= 9d1 ⇒ d1 = - and
⇒
⇒ ⇒
∴ a4h7 =
= 6 Create a free account to view solution
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