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If the numbers a, b, c, d and e form an A.P., then find the value of a – 4b + 6c – 4d + e.

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CBSE/Class 10/Mathematics Tuition/Class IX-X Tuition

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Let a = A-2D b = A-D c = A d = A+D e = A+2D a-4b+6c-4d+e = A-2D-4(A-D)+6A-4(A+D)+A+2D = 0

Maths Tutor

A, b , c , d and e are in AP let a = x b = x + d c = x + 2d d = x +3d e = x +4d put this in LHS LHS = a -4b +6c -4d +e = x -4(x +d ) +6(x +2d ) -4(x +3d) +x +4d ={ x -4x +6x -4x +x} +{ -4d +12d -12d +4d} = 0 + 0 = 0 = RHS


Ans is 0 .... U can take series as x-2y , x-y , x , x+y , x+2y in ap as a b c d e respectively and substitute


0. Take the AP series as a-2d,a-d,a,a+d,a+2d. Substitute and simplify.

Put a=x B= x+d C= x+2d D= x+3d E= x+4d Where d is the common difference solve the equation you will get zero

Science Teacher

the answer is 0 for detail answer see the attachment


Let Consider c=m since a, b, c, d, e are in AP Therefore, b=m-d (Here d is common difference) a=m-2d d=m+d e=m+2d Now, Putting the values of a, b, c, d and e in the final equation, we have =(m-2d)-4(m-d)+6m-4(m+d)+(m+2d) =m-2d-4m+4d+6m-4m-4d+m+2d =0

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