Question

The fourth term of a G.P. is eight times the 7 term. The fifth terms 3/16, then find its 12 term.​

Answer 1

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

• 4th term of GP = 8(7th term)
• 5th term of GP = 3/16 .
• Find 12th term of this GP ?

Concept used :---

$\bigstar\underline{\red{\mathbb{AP\:\green{SERIES}}}}\bigstar$

If in an A.P. series "a" be the first term and "d" be the common difference then ,

(1) The n'th term is given by the formula .

$\longrightarrow \red{\boxed{\sf a_n=a+(n-1)d}}$

(2)Sum of n number of terms ,

$\longrightarrow\green{\boxed{\sf S_n=\frac{n[2a+(n-1)d]}{2}}}$

━━━━━━━━━━━━━━━━━━━━━━━

$\bigstar\underline{\red{\mathbb{GP\:\pink{SERIES}}}}\bigstar$

If in an G.P. series "a" be the first term and "r" be the common ratio then ,

(1) The n'th term is given by the formula .

$\longrightarrow\pink{\boxed{\tt \: a_n=at{}^{n-1}}}$

(2)Sum of n number of terms ,

$\longrightarrow \purple{\boxed{\rm \: S_n=\frac{n[r{}^{n}-1]}{r-1}}}$

━━━━━━━━━━━━━━━━━━━━━━━

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

Let the First term be a and common ratio be r .

A/q,

→ a5 = 3/16

→ ar⁴ = 3/16 ------------ Equation (1)

Now,

→ a4 = 8(a7)

→ ar³ = 8(ar^6)

→ (ar^6/ar³) = 1/8

→ r³ = 1/8

→ r = 1/2 ---------------- Equation (2)

__________________________

Putting value of Equation (2) in Equation (1) we get,

→ a(1/2)⁴ = 3/16

→ a/16 = 3/16

→ a = 3

Now, so, our 12th term will be :---

→ ar^11 = 3*(1/2)^11 = 3/2^11 (Ans)

Hence, 12th term of GP will be 3/2^11 ....