Question

The second term of some arithmetic progression containing only whole numbers is 2 and the sum of the squares of third and fourth terms is less than 4. The first term of the progression is​

Answer 1

Answer:

First term of AP can be 3,2,1 or 0

Answer 2

Given:

The second term of some arithmetic progression containing only whole numbers is 2 and the sum of the squares of third and fourth terms is less than 4.

To Find:

The first term of the progression is​

Solution:

Let the first be a and the common difference of the arithmetic progression be 'd' so the given 2nd term 2 can be expressed as

a+d=2      -(1)

Now taking the other case as the sum of squares of the third and the fourth term is less than 4 we can then it as

$(a+2d)^2+(a+3d)^2<4$

Now substituting the value of equation 1 in this equation then we get

$(2+d)^2+(2+2d)^2<4\\5d^2+12d+4<0$

Solving the particular quadratic inequality we get

$d=\frac{-12\pm \sqrt{144-80} }{10}\\d=-2 or -0.4$

so d lies between -2 and -0.4 exclusive of ends

and it is said that an arithmetic progression is a whole number so we can only take -1 as the value of d

then

a+d=2

a=3

Hence, the first term of the AP will be 3.