Question

the 10th,4th and 1st term of an A.P are the three consecutive numbers of a G.P and the sum of the 1st 6 terms ;take 4 as the 1st term

Answer 1

Reiny did this. See the related questions below.

Answer 2

*Let the three terms of the AP be*

10th term = a + 9d

4th term = a + 3d

1st term = a

∴ (a + 9d), (a + 3d), and a

*If they also form a GP, then*

(a + 3d)/(a + 9d) = a/(a + 3d)

*Cross multiply*

(a + 3d)(a + 3d) = a(a + 9d)

a² + 3ad + 3ad + 9d² = a² + 9ad

a² + 6ad + 9d² = a² + 9ad

∴ a² + 6ad + 9d² - a² - 9ad = 0

∴ 9d² - 3ad = 0

∴ 3ad = 9d²

*But, we know that, a = 4*

∴ 3(4)d = 9d²

∴ 12d = 9d²

∴ 9d² - 12d = 0

∴ 3d(3d - 4) = 0

∴ 3d = 0 or 3d - 4 = 0

∴ d = 0/3 or 3d = 4

∴ d = 0, or d = 4/3

*Since common difference cannot be 0*

∴ d = 4/3

*so our three AP terms are:*

∴ (a + 9d), (a + 3d), and a

= 4 + 9(4/3) , 4 + 3(4/3) and 4

= 4 + 3(4) , 4 + 4 , and 4

= 16, 8 and 4 ,

*sure enough they also form a GP with, Common ratio, r = 1/2*

Now, a = 16

*Now, Sum of the first 6th terms of the G.P*

Sn = a[1 - rⁿ]/(1 - r)

S(6) = 16[1 - (1/2)^6]/(1 - 1/2)

S(6) = 16[1 - (1/64)]/(1/2)

S(6) = 16[(64 - 1)/64]/(1/2)

S(6) = (16 × 2)[(63)/64]

S(6) = (32)[(63)/64]

S(6) = 63/2

S(6) = 31.5

DONE !

Force × Distance = ISE = Dumb physics = Dumb ISE
*(I am a learner !!!)*