Question

If a1 , a2 , a3,a4 , a5 form an A.P then prove that a1-4a2+6a3-4a4+a5=0

Answer 1

I hope this answer would be helpful for you

Answer 2

$Let \:\pink {a} \:and \: \blue {d} \:are \: first \\teem \: and \: common \: differnce \: of \:an \:A.P$

$a_{1} = a$

$a_{2} = a + d$

$a_{3} = a + 2d$

$a_{4} = a + 3d$

$a_{5} = a + 5d$

$Now, LHS = \red { a_{1} - 4 \times a_{2} + 6a_{3} -4_{4} + a_{5}} \\= a - 4(a+d) +6(a+2d) - 4(a+3d) + (a+4d) \\= a - 4a - 4d + 6a +12d -4a-12d + a + 4d \\= 8a - 8a - 16d + 16d \\= 0 \\ = RHS$

$Hence\: proved$

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