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If $$a_1,$$ $$a_2,$$ $$.....a_n$$ are in $$H.P$$. then $${a_1}.{a_2} + {a_2}.{a_3} + {a_3}.{a_4} + ..... + {a_{n - 1}}.{a_n} = $$


Solution

Given, $$a_{1}, a_{2}, a_{3},.......a_{n}-H.P$$

$$\dfrac{1}{a_{1}}, \dfrac{1}{a_{2}}, ...... \dfrac{1}{a_{n}}-A.P$$

$$\dfrac{1}{a_{2}}-\dfrac{1}{a_{1}}=d, \dfrac{1}{a_{3}}-\dfrac{1}{a_{2}}=d, \dfrac{1}{a_{n}}-\dfrac{1}{a_{1}}=(n-1)\ d----- (A)$$

$$\dfrac{a_{1}-a_{2}}{a_{1}a{2}}=\dfrac{a_{2}-a_{3}}{a_{2}a_{3}}=\dfrac{a_{3}-a_{4}}{a_{3}a_{4}}=........=\dfrac{a_{n-1}-a_{n}}{a_{n-1}a_{n}}=d$$

$$a_{1}a_{2}=\dfrac{a_{1}-a_{2}}{d}.........a_{n-1}a_{n}=\dfrac{a_{n-1}-a_{n}}{d}$$

$$=a_{1}a_{2}+a_{2}+a_{3}+........a_{n-1}a_{n}$$

$$=\dfrac{a_{1}-a_{2}}{d}+\dfrac{a_{2}-a_{3}}{d}+...... \dfrac{a_{n-1}-a_{n}}{d}$$

$$=\dfrac{a_{1}-a_{n}}{d}$$

$$=(n-1)a\cdot a_{n}$$     ..................   From $$(A)$$

Mathematics

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