Question

In a G.P, the 5th term exceeds the 4th term by 24, and the 4th term exceeds the 3rd term by 8, find the common ratio and the first term.​

Answer 1

EXPLANATION.

The 5th term exceeds the 4th term by 24.

The 4th term exceeds the 3rd term by 8.

As we know that,

Formula of nth term of G.P.

⇒ Tₙ = arⁿ⁻¹.

The 5th term exceeds the 4th term by 24.

⇒ T₅ = T₄ + 24.

⇒ T₅ - T₄ = 24.

⇒ ar⁴ - ar³ = 24.

⇒ ar³(r - 1) = 24. - - - - - (1).

The 4th term exceeds the 3rd term by 8.

⇒ T₄ = T₃ + 8.

⇒ T₄ - T₃ = 8.

⇒ ar³ - ar² = 8.

⇒ ar²(r - 1) = 8. - - - - - (2).

Divide equation (1) by (2), we get.

⇒ ar³(r - 1) = 24. - - - - - (1).

⇒ ar²(r - 1) = 8. - - - - - (2).

We get,

⇒ r = 3.

Put the value of r = 3 in the equation (1), we get.

⇒ ar³(r - 1) = 24.

⇒ a(3)³(3 - 1) = 24.

⇒ a(27)(2) = 24.

⇒ 54a = 24.

⇒ a = 24/54.

⇒ a = 4/9.

First term = a = 4/9.

Common ratio = r = 3.

Answer 2

Answer:

Question :-

☯ In a G.P, the 5th term exceeds the 4th term by 24, and the 4th term exceeds the 3rd term by 8, find the common ratio and the first term.

Given :-

☯ In a G.P, the 5th term exceeds the 4th term by 24, and the 4th term exceeds the 3rd term by 8.

Find Out :-

☯ Find the common ratio and the first term.

Solution :-

☣ In first case :-

As we know that,

$\red{ \boxed{\sf{\bigstar\: T_n = ar^{n - 1}\: \bigstar}}}$

Given that :

✯ The 5th term exceeds the 4th term by 24.

$\implies \sf T_5 = T_4 + 24$

$\implies \sf ar^4 - ar^3 = 24$

$\implies \bf ar^3(r - 1) = 24\: -----\: (Equation\: No\: 1)$

☣ In second case :-

Given that :

✯The 4th term exceeds the 3rd term by 8.

$\implies \sf T_4 = T_3 + 8$

$\implies \sf ar^3 - ar^2 = 8$

$\implies \bf ar^2(r - 1) = 8\: -----\: (Equation\: No\: 2)$

By solving both the equation we get,

$\implies \sf \cancel{\dfrac{ar^3(r - 1)}{ar^2(r - 1)}} =\: \cancel{\dfrac{24}{8}}$

$\implies$ ${\small{\bold{\purple{\underline{r = 3}}}}}$

By put the value of r = 3 in the equation (2), we get.

$\implies$ ar²(r - 1) = 8

$\implies$ a(3)²(3 - 1) = 8

$\implies$ a(3)²(3 - 1) = 8

$\implies$ a(9)(2) = 8

$\implies$ a(18) = 8

$\implies$ 18a = 8

$\implies$ a = $\sf \cancel{\dfrac{8}{18}}$

$\implies$ ${\small{\bold{\purple{\underline{a =\: \dfrac{4}{9}}}}}}$

Henceforth, common ratio is 3 and first term is 4/9.

$\purple{\rule{45pt}{7pt}}\red{\rule{45pt}{7pt}}\pink{\rule{45pt}{7pt}}\blue{\rule{45pt}{7pt}}$