Question

there are five consecutive terms in an AP the product of first and last term is 189 if the first term is less than 4th term find the AP

Answer 1

Correct question : there are five consecutive terms in an AP the product of first and last term is 189 if the first term is 9 less than 4th term find the AP

Solution :

For AP, the nth term can be written as:

Tₙ = a + (n-1)d

For the 4th term,

T₄ = a + (4-1)d

=> T₄ - a = 3d

=> 9 = 3d

=> d = 3

Similarly

T₅ = a + (5-1)3

multiplying a on both sides,

T₅ x a = a(a + 12)

=> 189 = a² + 12a

=> a² + 12a - 189 = 0

=> (a+21)(a-9) = 0

=> a = 9

Hence the AP is (9, 12, 15, 18, 21)

Answer 2

Answer:

_____________

A.P. will be :

{ 9, 12, 15, 18, 21 }

OR { -21, -18, -15, -12, -9 }

SOLUTION :

_______________

Let, 5 consecutive terms of A.P be :

a, a+d, a+2d, a+3d, a+4d

As given that :

Product of first and last term = 189

a×(a+4d)=189............(1)

Also, given that :

$t_{1} = t_{4} - 9 \\ \\ = > a = a + 3d - 9 \\ \\ = > 3d = 9 \\ \\ = > d = \frac{9}{3} = 3$

On putting the value of 'd' in equation (1)

a(a+4×3) = 189

=> a²+12a-189 = 0

=> a²+21a-9a-189 = 0

=> a(a+21)-9(a+21) = 0

=> (a+21)(a-9) = 0

So, a = 9 and -21

When, the value of a = 9

Then, required A.P. will be :

9, 12, 15, 18, 21

And when, the value of a=-21

Then, required value of A.P. will be :

-21, -18, -15, -12, -9