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

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

$\displaystyle\sum_{k=1}^2(8+21(k-1))=37\neq123$

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

$\displaystyle\sum_{k=1}^{22}(8+1(k-1))=407\neq123$

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

$\displaystyle\sum_{k=1}^4(8+7(k-1))=74\neq123$

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

$\displaystyle\sum_{k=1}^8(8+3(k-1))=148\neq123$

Answer 2

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 //