Question

find first four terms of sequence of which sum to n terms is 1/2n (7n-1)

Answer 1

Given, $\bold{S_n=\frac{1}{2}n(7n-1)}$

We know, one important concept,
$\bold{T_n=S_n-S_{n-1}}$ use it here,
$\bold{S_n=\frac{1}{2}n(7n-1)}\\\\\bold{S_{n-1}=\frac{1}{2}(n-1)(7n-8)}$
So, $\bold{T_n=\frac{1}{2}n(7n-1)-\frac{1}{2}(n-1)(7n-8)}\\\\=\bold{\frac{1}{2}[7n^2-n-7n^2+15n-8]}\\\\=\bold{\frac{1}{2}(14n-8)}\\\\=\bold{7n-4}$

Now, Tn = 7n - 4
First term , n = 1 ⇒T₁ = 7 × 1 - 4 = 3
second term , n = 2 ⇒, T₂ = 7 × 2 - 4 = 10
3rd term , n = 3 ⇒T₃ = 7 × 3 - 4 = 17
4th term , n = 4 ⇒T₄ = 7 × 4 - 4 = 24