Question

if tn is the nth term of an A.P. then the value of tn+1 – tn-1 is​

Answer 1

The sum of n terms is also equal to the formula where l is the last term

. Tn = Sn - Sn-1 , where Tn = nth term.

When three quantities are in AP, the middle one is called as the arithmetic mean of the other two.

If a, b and c are three terms in AP then b = (a+c)/2.

The general form of an Arithmetic Progression is a, a + d, a + 2d, a + 3d and so on. Thus nth term of an AP series is Tn = a + (n - 1) d, where Tn = nth term and a = first term. Here d = common difference = Tn - Tn-1.

Answer 2

Answer:

$t_{n+1} - t{n-1}$  =  twice the common difference  of the AP

Step-by-step explanation:

Given,

The nth tern of an AP = tₙ

To find,

The value of  $t_{n+1} - t{n-1}$

Recall the formula

The nth term of an AP is given by

tₙ = a+(n-1)d, where a is the first term and 'd' is the common difference of the AP

Solution

Since tₙ = a+(n-1)d

$t_{n+1}$ = a+ (n+1-1) d

=a+nd

$t_{n+1}$  = a+nd

$t_{n-1} =$ a + (n-1-1)d

= a + (n-2)d

$t_{n-1} =$ a+(n-2)d

$t_{n+1} - t{n-1}$ = a+nd - ( a + (n-2)d)

= a+nd -a -(n-2)d

= a+nd -a -nd +2d

= 2d

= 2× common difference

∴ $t_{n+1} - t{n-1}$  =  twice the common difference  of the AP

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