Question

in a AP the first term is 8,nth term is 33 and sum to first n terms is 123.Find n and common difference​

Answer 1

a = 8

Un = L = 33 ( nth term = last term)

Sn = 123

Sn = n/2 ( a + L)

123 = n/2 ( 8 + 33 )

123 = n/2 (41)

246 = n (41)

n = 246/41

n = 6

Un = a + (n-1)d

33 = 8 + (6-1)d

33 = 8 + 5d

25 = 5d

d = 25/5

d = 5 //

Answer 2

I believe there is no such AP...

Recursively, this sequence is supposed to be given by

`\begin{cases}a_1=8\\a_k=a_(k-1)+d&\text{for }k>1\end{cases}`

so that

`a_k=a_(k-1)+d=a_(k-2)+2d=\cdots=a_1+(k-1)d`

`a_n=a_1+(n-1)d`

`33=8+(n-1)d`

`21=(n-1)d`

`n` has to be an integer, which means there are 4 possible cases.

Case 1:
`n-1=1` and
`d=21`. But

`\displaystyle\sum_(k=1)^2(8+21(k-1))=37\\eq123`

Case 2:
`n-1=21` and
`d=1`. But

`\displaystyle\sum_(k=1)^(22)(8+1(k-1))=407\\eq123`

Case 3:
`n-1=3` and
`d=7`. But

`\displaystyle\sum_(k=1)^4(8+7(k-1))=74\\eq123`

Case 4:
`n-1=7` and
`d=3`. But

`\displaystyle\sum_(k=1)^8(8+3(k-1))=148\\eq123`