Question

in an arithmetic progression of 16 distinct terms with a1= 16 , the sum is equal to square of the last term. The common difference of the A.P. is :

Answer 1

Answer:

common difference of AP is 8/5

Answer 2

Given:

in an arithmetic progression of 16 distinct terms with a1= 16, the sum is equal to the square of the last term.

To Find:

The common difference of the A.P. is

Solution:

An arithmetic progression is a progression in which every consecutive term differs by a common difference which is denoted by d and the first term of an AP is denoted by a.

The nth term of an AP can be expressed as,

$T_n=a+(n-1)d$

And the sum of n terms of an AP is expressed as,

$S_n=\frac{n}{2}(2a+(n-1)d)$

Now it is given that the first term of the AP is 16, and the total number of terms is 16, so we are available with,

a=16

n=16

Now putting the values in the equation formed by the condition that  the sum is equal to the square of the last term,

$\frac{16}{2}(2*16+(16-1)d) =(16+15d)^2\\\\256+120d=256+225d^2+480d\\\\d(225d+360)=0\\\\d=\frac{-360}{225} \\\\d=\frac{-8}{5}$

Hence, the common difference of the AP is -8/5.