Question

Let $a_1$,$a_2$,............$a_{10}$ be in AP and $h_1$, $h_2$..............$h_{10}$ be in H.P. If $a_1$ = $h_1$ = 2 and $a_{10}$ = $h_{10}$ = 3, then $a_4$$h_7$ is

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Answer 1

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

$a_1$,$a_2$,............$a_{10}$ be in AP

$a_1$ + 9d = $a_{10}$

$\Rightarrow$ 2 + 9d = 3

$\Rightarrow$ d = $\frac{1}{9}$

$a_4$ = $a_1$ + (4-1)d

= 2 + 3 $\times$ $\frac{1}{9}$

$\Rightarrow$ $a_4$ = $\frac{7}{3}$

$h_1$, $h_2$..............$h_{10}$ are in H.P

$\Rightarrow$ $\frac{1}{h_1}$,$\frac{1}{h_2}$,$\frac{1}{h_3}$,...................$\frac{1}{h_{10}}$ are in A.P.

let the common difference of this A.P be d

$\frac{1}{h_{10}}$ = $\frac{1}{h_1}$ +9.d

$\frac{1}{3}$ = $\frac{1}{2}$ + 9.d

9.d = $\frac{1}{3}$ - $\frac{1}{2}$

= -$\frac{1}{6}$

d = $\frac{-1}{54}$

$\frac{1}{h_7}$ = $\frac{1}{h_1}$ + (7 - 1)d

= $\frac{1}{2}$ + 6 $\times$ $\frac{-1}{54}$

= $\frac{1}{2}$ - $\frac{1}{9}$

= $\frac{7}{18}$

$\Rightarrow$ $h_7$ = $\frac{18}{7}$

$\therefore$ $a_4$$h_7$ = $\frac{7}{3}$ $\times$ $\frac{18}{7}$

= 6