# for a given arithmetic sequence, un=M and um=n find (a) the common difference (b) its first term

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um = n

un = m

a) the common difference (r) is

let u1, u2, u3, ...., um,...un be an arthimatical sequence:

u1 = u1

u2= u1 + r

u3 = u1 + 2r

.....

um = u1 + (m-1)r = n ......(1)

un = u1 + (n-1)r = m........(2)

Subtract (2) from (1):

==> (m-1)r - (n-1)r = n - m

==> (m-1-n + 1) r = (n-m)

==> (m-n) r = (n-m)

Divide by (m-n)

==> r = (n-m)/(m-n) = -1

**Then the common differecnse is r = -1**

b) the first term:

We know that:

un = u1 + (n-1)r = m

==> u1 + (n-1)*-1 = m

==> u1 + 1 - n = m

==> **u1 = m + n -1**

**Then the first terms is :**

**u1 = m + n -1**

The given arithmetic progression has the nth term un = M and the mthterm um = n.

To find the (i)common difference d and the first term.

Solution:

The relation between the first term a1 and the rth term ar and the common diference d for an arithmetic progression(AP) is given by:

ar = a1+(r-1)d.

Now by the above relation , the formula for the given AP, nth and mth terms could be rewriten as below:

ur = u1+(r-1)d becomes relation.

um = u1+(n-1)d = N ......(1)

um = u1+(m-1)d = m.....(2)

Now solve for u1 and d in terms of n,m and N

(1) -(2) gives: u1+(n-1)d - u1-(m-1)d = N-m

(n-m)d = N-m

**d = (N-m)/(n-m)** is the common difference.

The first term u1 is got by substituting the value of d in eq (1)(any one of the 2equations (1) or (2)).

u1+(n-1)d = N.

u1 + (n-1)(N-m)/(n-m) = N

U1 = N- (n-1)(N-m)/(n-m)

u1 = {N(n-m)- (n-1)(N-m)}/(n-m)

u1 = {Nn -Nm - nN +nm+N -m}/(n-m)

u1 = {nm -Nm +N-m}/(n-m)

u1 = {n(m-N) - 1(m-N)}/(n-m)

**u1 = (m-N)(n-1)/(n-m) is the 1st term.**