Question

an=(-1)n-1 ×5n+1
class 11 maths
ch 9
ex 9.1
question no. 5
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Answer 1

We need to find first five terms i.e.

a¹, a², a³, a⁴, a⁵

Let a = (-1)-150---> 1

Putting n = 1 in (1)

a, =(-1) 151 1 = (-1)" x 5² = 1 x 5² = 5x5 = 25

Putting n = 2 in (1)

= -1(125) a_{2}

= (- 1) ^ (2 - 1) * 5 ^ (2 + 1)

= (-1) ¹ x 5³

=(-1)×(5 x 5 x 5)

= -125

Putting n = 3 in (1)

a =(-1)³-1 5³+1

= (-1)² x 54 = 1 x 54

= (5 x 5 x 5 x 5)

= 625

Putting n = 4 in (1)

a =(-1) 4-1 54+1

= (-1) ³ (5)

= (-1)(5)

= (-1) (5 x 5 x 5 x 5 x 5)

= -1(3125)

= -3125

Putting n = 5 in (1)

a_{5} = (- 1) ^ (5 - 1) * 5 ^ (5 + 1)

= (-1)" (5) = 1x56

=1(5x5x 5x5x 5* 5)

= 15625

Hence first five terms are 25, -125, 625,-3125, and 15625.

Answer 2

First we write the first five terms of the sequences whose nth term is$\sf a_n =(-1)^{n-1}× 5^{n+1}$

Substituting n =1,2,3,4,5,, We obtain

$\sf{a_1=(-1)^{1-1}×5^{1+1}}$

$\sf{a_1=(-1)^0 ×25}$

$\sf{a_1=1×25}$

$\sf{a_1=25}$

__________

$\sf{a_2=(-1)^{2-1}×5^{2+1}}$

$\sf{a_2= - 1×5^{3}}$

$\sf{a_2= - 125}$

__________

$\sf{a_3=(-1)^{3-1}×5^{3+1}}$

$\sf{a_3=(-1)^{2}×5^{4}}$

$\sf{a_3=1×625}$

$\sf{a_3=625}$

__________

$\sf{a_4=(-1)^{4-1}×5^{4+1}}$

$\sf{a_4=(-1)^{3}×5^{5}}$

$\sf{a_4= - 1×3125}$

$\sf{a_4= -3125}$

_________

$\sf{a_5=(-1)^{5-1}×5^{5+1}}$

$\sf{a_5=(-1)^{4}×5^{6}}$

$\sf{a_5= - 1×15625}$

$\sf{a_5= -15625}$

• Therefore , the required terms are 25 , -125 , 625 , -3125 , and 15625 .