Question

-1/3,0,1/3,2/3 are sequencesin A.P​

Answer 1

Step-by-step explanation:

$\huge{\pink{\boxed{\green{\underline{\red{\sf{SOLUTION-}}}}}}}$

$\: \: \: \: \: \: \: \: \: \: \: { \blue{ to \: show}} \\ { \pink{ \boxed{ \red{ \boxed{ \frac{ - 1}{3},0, \frac{1}{3},\frac{2}{3} \: }are \: in \: A.P}}}}$

☆According to given question

☆ We need to prove that the given sequence is in A.P

☆ PROOF:

$\to a2 - a1 = a3 - a2 \\ \: \: \: \: \: \: \: \: <strong>L</strong><strong>H</strong><strong>S</strong> \\ \to 0 - \frac{ - 1}{3} \\ \to \frac{1}{3} \\ \: \: \: \: \: \: <strong>R</strong><strong>H</strong><strong>S</strong>\\ \to \frac{1}{3} - 0 \\ \to \frac{1}{3} \\ \: \: \: \: \: \: <strong>L</strong><strong>H</strong><strong>S</strong><strong>=</strong><strong>R</strong><strong>H</strong><strong>S</strong>$

${\pink{\boxed{\green{\sf{So,\:given\:sequence\:in\:an\:A.P}}}}}$

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Answer 2

Step-by-step explanation:

$- \frac{ 1}{3} , \: 0,\: \frac{1}{3}, \: \frac{2}{3} \\ \\ a = - \frac{1}{3} \\ t2 = 0 \\ t3 = \frac{1}{3} \\ \\ d = t2 - t1 = 0 - ( - \frac{1}{3} ) = \frac{1}{3} \\ d = t3 - t2 = \frac{2}{3} - \frac{1}{3} = \frac{1}{3} \\ \\ as \: common \: difference \: is \: same \: \\ so \: they \: are \: in \: A. P$