Question

Find the first four terms of a sequence of which sum to $n$ terms is $\frac{1}{2}n(7n-1)$

Answer 1

Thegiven sum is $S_n=\frac{n}{2}(7n-1)$, therefore,

$S_1=\frac{1}{2}\left[\right(7\times 1)-1]=\frac{1}{2}(7-1)=\frac{6}{2}=3$

$S_2=\frac{2}{2}\left[\right(7\times 2)-1]=(14-1)=13$

$S_3=\frac{3}{2}\left[\right(7\times 3)-1]=\frac{3}{2}(21-1)=\frac{60}{2}=30$

$S_4=\frac{4}{2}\left[\right(7\times 4)-1]=2(28-1)=2\times 27=54$

Now, the terms of the sequence can be obtained as follows:

$T_1=S_1=3$

$T_2=S_2-S_1=13-3=10$

$T_3=S_3-S_2=30-13=17$

$T_4=S_4-S_3=54-30=24$

Hence, the first four terms of the sequence are $3,10,17,24$.