Question

ii. For a G.P.Sum of first 3 terms is 125 and sum
of next 3 terms is 27. Find the value of r.
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Answer 1

Answer:

r=3/5

Step-by-step explanation:

Let the GP be a,ar,ar^2....ar5

a+ar+ar^2=125

a(1+r+r^2)=125........(1)

For next 3 terms,

ar^3+ar^4+ar^5=27

ar^3(1+r+r^2)=27........(2)

Keeping (1) in (2),

r^3=27/125

r=3/5

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Answer 2

Answer:

$\bigstar{\bold{Value\:of\:r=\dfrac{3}{5} }}$

Step-by-step explanation:

$\Large{\underline{\bf{Given:}}}$

• Sum of first 3 terms = 125
• Sum of next three terms = 27

$\Large{\underline{\bf{To\:Find:}}}$

• The value of r

$\Large{\underline{\bf{Solution:}}}$

➔ Let us assume the first term of the A.P as a

➔ Hence,

    The second term = ar

    The third term = ar²

➔ By given,

    Sum of first 3 terms = 125

    a + ar + ar² = 125

➔ Taking a as common,

    a (1 + r + r²) = 125

    1 + r + r² = 125/a -------(1)

➔ Now,

   Fourth term = ar³

   Fifth term = ar⁴

   Sixth term = ar⁵

➔ Also by given,

    Sum of next 3 terms = 27

    ar³ + ar⁴ + ar⁵ = 27

➔ Taking ar³ as common,

    ar³ (1 + r + r²) = 27

    1 + r + r² = 27/ar³ -------(2)

➔ From equations 1 and 2, LHS are equal, hence RHS must also be equal.

    125/a = 27/ar³

➔ Cancelling a on both sides,

     125 = 27/r³

     125 r³ = 27

     r = 27/125

     r = ∛ (27/125)

    r = 3/5

➔ Hence the value of r is 3/5.

    $\boxed{\bold{Value\:of\:r=\dfrac{3}{5} }}$