Question

Question 8 The sum of some terms of G.P. is 315 whose first term and the common ratio are 5 and 2, respectively. Find the last term and the number of terms.

Class X1 - Maths -Sequences and Series Page 199

Answer 1

Hey there!

Let the sum of n terms of the G.P. be 315.

$Sn \: = \: \frac{a(r^{n} \: - 1) }{r \: - \: 1}$

We know that,

It is given that the first term a is 5 and common ratio r is 2.

$315 = \frac{5(2^n - 1)}{2 - 1}$

${2}^{n} - 1 = 63$

$n = 6$

∴ Last term of the G.P = 6th term

$= {ar}^{6 - 1 } = (5)(2)^{5} = 160$

Therefore, the last term of the G.P. is 160.

Answer 2

Let GP of n terms is a, ar , ar², ar³,......arⁿ⁻¹
Here,
first term ( a ) = 5
common ratio ( r ) = 2

Let Sum of n terms ( Sₙ ) = 315
use formula,
Sₙ = a (rⁿ - 1)/(r - 1)
Therefore,
315 = 5( 2ⁿ - 1)/(2 - 1)
63 = 2ⁿ - 1
64 = 2ⁿ
2⁶ = 2ⁿ
n = 6
hence, number of terms = 6

now,
nth term = Tₙ = arⁿ⁻¹
= 5 × (2)⁶⁻¹
= 5 × 2⁵
= 5 × 32
= 160