Question

If -5/7, a, 2 are consecutive terms in an Arithemetic Progression, then the value of ‘a’ is *​class 10th.....

Answer 1

Given :

• ${\dfrac{-5}{7}}$ , a, 2 are consecutive terms of an Arithmetic progression

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To find :

• The value of 'a'

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Concept :

In an arithmetic progession the Common difference ( d ) = ${t}_{2} - {t}_{1} = {t}_{3} - {t}_{2}$

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Solution :

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${t}_{1} = {\dfrac{-5}{7}}$

${t}_{2} = a$

${t}_{3} = 2$

Common difference ( d ) = ${t}_{2} - {t}_{1} = {t}_{3} - {t}_{2}$

$\implies {a-({\dfrac{-5}{7}}) = 2-a}$

$\implies {a+{\dfrac{5}{7}} = 2-a}$

$\implies {{\dfrac{7a+5}{7}} = 2-a}$

$\implies {7a+5 = 7 (2-a)}$

$\implies {7a+5 = 14-7a}$

$\implies {7a+7a = 14-5}$

$\implies {14a =9 }$

$\implies{\boxed{a={\dfrac{9}{14}}}}$

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We can also verify the value of a by substituting the value of a in the equation ...

Verification :

$\implies{{\dfrac {9}{14}}-({\dfrac{-5}{7}}) = 2-{\dfrac{9}{14}}}$

$\implies{{\dfrac {9}{14}}+{\dfrac{5}{7}}) = 2-{\dfrac{9}{14}}}$

$\implies {{\dfrac{9+10}{14}} = {\dfrac{28-9}{14}} }$

$\implies {{\dfrac{19}{14}} = {\dfrac{19}{14}} }$

$\therefore{LHS = RHS }$