Question

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

Answer 1

Answer:

Step-by-step explanation:

Let the first term be ′a′the common ratio be ′r′

Given ar2=12→(1)

ar5=96→(2)

Divide equation(2)by(1)

r3=8

Hence r=2

Substitute in equation (1) to get a=3

Hence sum of nine terms S9=3(29−1)/(2−1)=1533

Answer 2

Answer: The A.P. will be 9,12,15,18,21 and -21,-18,-15,-12,-9

Step-by-step explanation:

Since we have given that

The product of first and the last term is 189.

so, it becomes,

$a_1\times a_5=189\\\\a(a+(5-1)d)=189\\\\a(a+4d)=189-------------(1)$

Now, we have also given that

The first term is 9 less than the 4th term .

$a_4-a_1=9\\\\a+3d-a=9\\\\3d=9\\\\d=\frac{9}{3}\\\\d=3$

So, put the value of d in Eq.(1), we get that

$a(a+4\times 3)=189\\\\a^2+12a-189=0\\\\a^2+21a-9a-189=0\\\\a(a+21)-9(a+21)=0\\\\(a+21)(a-9)=0\\\\a=-21,9$

Hence, the A.P. will be

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