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Question

Let a1,a2,............a10 be in AP and h1, h2..............h10 be in H.P. If a1 = h1 = 2 and a10 = h10 = 3, then a4h7 is


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Solution

We are given the first and last term of the A.P and H.P.

We can use it to find the common difference of the A.P

and the common difference of the A.P corresponding to the given H.P.

Once we have common difference and first,

we can find any term of the A.P and H.P

a1,a2,............a10 be in AP

a1 + 9d = a10

2 + 9d = 3

d = 19

a4 = a1 + (4-1)d

= 2 + 3 × 19

a4 = 73

h1, h2..............h10 are in H.P

1h1,1h2,1h3,...................1h10 are in A.P.

let the common difference of this A.P be d

1h10 = 1h1 +9.d

13 = 12 + 9.d

9.d = 13 - 12

= -16

d = 154

1h7 = 1h1 + (7 - 1)d

= 12 + 6 × 154

= 12 - 19

= 718

h7 = 187

a4h7 = 73 × 187

= 6


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