If x_1,x_2,.,x_n and 1h_1,1h_2,..1h_n are two A.P.s such that x_3 = h_2 = 8 and x_8 = h_7 = 20, then x_5 h_10 equals

(a) : Let d_1 and 1 / d_2 be the common difference of A.P. x_1, x_2, , x_n and 1h_1, 1h_2, 1h_3, , 1h_n respectively., x_3=8 ; x_8=20 x_1+2 d_1=8 (i) and x_1+7 d_1=20 solving (i) and (ii), we get d_1=125 and x_1=165; h_2=8 and h_7=20 1h_2=18 and 1h_7=120 1h_1+1d_2=18 1h_1+6d_2=120 solving (iii) and (iv), we get 1d_2=-3200, 1h_1=28200, x_5=x_1+4 d_1

If x_{1},x_{2},.,x_{n} and \frac{1}{h_{1}},\frac{1}{h_{2}},..\frac{1}{h_{n}} are two A.P.s such that x_{3} = h_{2} = 8 and x_{8} = h_{7} = 20, then x_{5} \cdot h_{10} equals: Sequence & Series (Mathematics)

(a) : Let

d_1

and

1 / d_2

be the common difference of A.P.

x_1, x_2, \ldots, x_n

and

\frac{1}{h_1}, \frac{1}{h_2}, \frac{1}{h_3}, \ldots, \frac{1}{h_n}

respectively.\nGiven,

x_3=8 ; x_8=20

\n\Rightarrow x_1+2 d_1=8 \ldots \text { (i) and } x_1+7 d_1=20\n

\n \nOn solving (i) and (ii), we get

d_1=\frac{12}{5}

and

x_1=\frac{16}{5}

\nSimilarly;

h_2=8

and

h_7=20

\n\Rightarrow \frac{1}{h_2}=\frac{1}{8} \text { and } \frac{1}{h_7}=\frac{1}{20} \Rightarrow \frac{1}{h_1}+\frac{1}{d_2}=\frac{1}{8}\n

\nand

\frac{1}{h_1}+\frac{6}{d_2}=\frac{1}{20}

\nOn solving (iii) and (iv), we get

\frac{1}{d_2}=\frac{-3}{200}, \frac{1}{h_1}=\frac{28}{200}

\nNow,

x_5=x_1+4 d_1=\frac{16}{5}+\frac{48}{5}=\frac{64}{5}

\nAlso,

\frac{1}{h_{10}}=\frac{1}{h_1}+\frac{9}{d_2}=\frac{28}{200}-\frac{27}{200}=\frac{1}{200}

\n\therefore x_5 \cdot h_{10}=\frac{64}{5} \times 200=2560\n

Explanation

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