Question

The sum of the first n terms of an AP, 2,6,10, is 882. Find the value of n

Answer 1

\[
a_i= 4(i-1) +2 \text { for } \quad i=1,2,3,\cdots

\]

Answer 2

\[
\sum_{i=1}^n 4(i-1) + 2 = 2n + 4( 1+2+\cdots n-1)= 288\\
2n + 4 n(n-1)/2 = 288
2 n^2 =288\\
n^2=144\\
n=12
\]

Answer 3

\[
\sum_ {i = 1}^n 4 (i - 1) + 2 =
2 n + 4 (1 + 2 + \cdots n - 1) = 288 \\
\]

Answer 4

\[2 n + 4 n (n - 1)/2 =
288\implies 2 n^2 = 288\implies n^2 = 144\implies n = 12\]
\]