Question

5 terms of are AP the products of first and last term is 28 if the 1st term is 6 less than its third term . Then find AP?

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Answer:

Number of terms = 5

We are given that  the products of first and last term is 28

So, $a_1 \times a_5=28$

Formula of nth term = $a_n=a+(n-1)d$

Substitute n =5

So, $a_5=a+(5-1)d$

$a_5=a+4d$

So,  $a(a+4d)=28$ ---1

Now we are given that the 1st term is 6 less than its third term .

So, $a_1+6 = a_3$

$a+6 = a+2d$

$6 = 2d$

$d=3$

Substitute the value in 1

$a(a+4\times 3)=28$

$a(a+12)=28$

$a^2+12a-28=0$

$a^2+14a-2a-28=0$

$a(a+14)-2(a+14)=0$

$(a-2)(a+14)=0$

$a=2,-14$

when a =2

AP = 2,2+3,2+3+3,2+3+3+3,2+3+3+3+3

AP = 2,5,8,11,14

when a = -14

AP = -14,-14+3,-14+3+3,-14+3+3+3,-14+3+3+3+3

AP = -14,-11,-8,-5,-2