न —— >> । — 3का ang ५. #-९िशा: 5. हिल �

1.	न —— >> । — 3का ang ५. #-९िशा: 5. हिल पी o sB bl e
Answer» Let x-1, x-4, x-7 are in ap. So find the value of x. To check whether these terms are in AP (x-4) - (x-1) = (x-7) - (x-4) = d = common difference => x - 4 -x + 1 = x - 7 -x + 4 => -3 = -3 So these terms are in AP for any given value of x with common difference = -3 If these terms are in GP then (x-4)/(x-1) = (x-7)/(x-4) = r = common Ratio => (x-4)² = (x-7)(x-1) => x² + 16 - 8x = x² -8x + 7 => 16 = 7 which is not possible hence these terms can not be in GP for any value of x Like my answer if you find it useful!

न —— >> । — 3का ang ५. #-९िशा: 5. हिल पी o sB bl e

Answer»

Let x-1, x-4, x-7 are in ap. So find the value of x.

To check whether these terms are in AP

(x-4) - (x-1) = (x-7) - (x-4) = d = common difference

=> x - 4 -x + 1 = x - 7 -x + 4

=> -3 = -3

So these terms are in AP for any given value of x

with common difference = -3

If these terms are in GP then

(x-4)/(x-1) = (x-7)/(x-4) = r = common Ratio

=> (x-4)² = (x-7)(x-1)

=> x² + 16 - 8x = x² -8x + 7

=> 16 = 7

which is not possible

hence these terms can not be in GP for any value of x

Like my answer if you find it useful!