If x_1,x_2, ... .,x_n andn frac1h_1,frac1h_2, ... ..frac1h_n are twon A.P.s such that x_3 = h_2 = 8 and x_8 = h_7 = 20, thenn 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 (1 / h_1), (1 / h_2), (1 / h_3), ... , (1 / h_n) respectively.nGiven, x_3=8 ; x_8=20nn => x_1+2 d_1=8 ... (i) and x_1+7 d_1=20nn nOn solving (i) and (ii), we get d_1=(12 / 5) and x_1=(16 / 5)nSimilarly; h_2=8 and h_7=20nn => (1 / h_2)=(1 / 8) and (1 / h_7)=(1 / 20) => (1 / h_1)+(1 / d_2)=(1 / 8)nnand (1 / h_1)+(6 / d_2)=(1 / 20)nOn solving (iii) and (iv), we get (1 / d_2)=(-3 / 200), (1 / h_1)=(28 / 200)nNow, x_5=x_1+4 d_1=(16 / 5)+(48 / 5)=(64 / 5)nAlso, frac1h_10=(1 / h_1)+(9 / d_2)=(28 / 200)-(27 / 200)=(1 / 200)nnTherefore x_5 * h_10=(64 / 5) x 200=2560n

If x_{1},x_{2},.,x_{n} and \frac{1}{h_{1}},\frac{1}{h... | 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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