Math, asked by nikhil9523, 1 year ago

a particle initially at rest starts moving in a straight line with an acceleration a=6t+4m/s².the distance covered by it in 3 sec is​

Answers

Answered by primishra
1

Step-by-step explanation:

» Four numbers are in AP whose sum is 40.

• Let four numbers be (a - 3d), (a - d), (a + d), (a + 3d)

A.T.Q.

=> a - 3d + a - d + a + d + a + 3d = 40

=> 4a = 40

=> a = 40/4

=> a = 10 _______ (eq 1)

__________________________________

» The ratio of the product of first and fourth term is 2:3.

\dfrac{a \: - \: 3d}{a \: + \: 3d} \: = \: \dfrac{2}{3}

a+3d

a−3d

=

3

2

Cross-multiply them..

=> 3(a - 3d) = 2(a + 3d)

=> 3a - 9d = 2a + 6d

=> 3(10) - 9d = 2(10) + 6d [From (eq 1)]

=> 30 - 9d = 20 + 6d

=> 30 - 20 = 6d + 9d

=> 10 = 15d

=> d = 10/15

=> d = 2/3 ________ (eq 2)

_______________________________

» We have to find the numbers. (Means AP.)

• a - 3d = 10 - 3(2/3)

=> 10 - 2 = 8

• a - d = 10 - 2/3

=> (30 - 2)/3 = 28/3

• a + d = 10 + 2/3

=> (30 + 2)/3 = 32/3

• a + 3d = 10 + 3(2/3)

=> 10 + 2 = 12

_______________________________

\textbf{Number are : 8, 28/3, 32/3 and 12}Number are : 8, 28/3, 32/3 and 12

__________ [ANSWER]

________________________________

✡ From above calculations we get numbers 8, 28/3, 32/3 and 12

Add all the numbers.

=> 8 + 28/3 + 32/3 + 12

=> (24 + 28 + 32 + 28)3

=> 120/3

=> 40

_______ [HENCE VERIFIED]

______________________________

Similar questions