Question

in an ap an=2n-1 and an+1=2+n It's common difference is​

Answer 1

Solution :

Let us assume that the ap has the starting term as a and the common difference d .

a_n = 2n - 1

a_(n+1) = 2 + n

a + (n-1)d = 2n - 1

&

a + nd = 2 + n

Subtracting the first equation from the second equation

> [ a + nd ] - [ a + (n-1)d ] = 2 + n - (2n -1)

> a + nd - a - (n-1)d = 2 + n - 2n + 1

> nd - nd + d = 2-n + 1

> d = 3 - n

Now

a + (n-1)d = 2n - 1

> a + (n-1)(3-n) = 2n - 1

> a + 3n - n² - 3 + n = 2n - 1

> a = n² - 2n + 2

Answer :

a = n² - 2n + 2 & d = 3 - n .

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Answer 2

Step-by-step explanation:

Solution :

Let us assume that the ap has the starting term as a and the common difference d .

a_n = 2n - 1

a_(n+1) = 2 + n

a + (n-1)d = 2n - 1

&

a + nd = 2 + n

Subtracting the first equation from the second equation

> [ a + nd ] - [ a + (n-1)d ] = 2 + n - (2n -1)

> a + nd - a - (n-1)d = 2 + n - 2n + 1

> nd - nd + d = 2-n + 1

> d = 3 - n

Now

a + (n-1)d = 2n - 1

> a + (n-1)(3-n) = 2n - 1

> a + 3n - n² - 3 + n = 2n - 1

> a = n² - 2n + 2

Answer :

a = n² - 2n + 2 & d = 3 - n .

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