Question

The Sum of three consecutive terms of an A.p is 33 and their product is 792.The least one of them is?

Answer 1

Let us assume that the 3 terms are a-d, a, a+d.

their sum = 33
a+d+a+a-d = 33
3a = 33
a = 11

their products =792
(a-d)×a×(a+d)= 792

a²-d² =792/a= 792/11
a²-d² = 792/11=72
d²= a² - 72
= 11²-72
= 121-72= 49
d²=49
d = 7
The terms in AP are 4,11,18.
The least term is 4.

Answer 2

Answer:

The least number among them is 4.

Step-by-step explanation:

Given : The Sum of three consecutive terms of an A.p is 33 and their product is 792.

To find : The least one of them is?

Solution :

Let the arithmetic progression series in the form,

(a-d) , a, (a+d)

We have given,

The Sum of three consecutive terms of an A.p is 33.

i.e. $a-d+a+a+d=33$

$3a=33$

$a=\frac{33}{3}$

$a=11$

Their product is 792

i.e. $(a-d)\times a\times (a+d)=792$

$a(a^2-d^2)=792$

$11(11^2-d^2)=792$

$121-d^2=72$

$d^2=49$

$d=\pm7$

So, The series form when a=11 and d=7,

(11-7) , 11, (11+7) = 4,11,18

The series form when a=11 and d=-7,

(11-(-7)) , 11, (11-7) = 18,11,4

Therefore, The least number among them is 4.