Question

If the sum of the first $15$ terms of the series $3+7+14+24+37+.......$ is $15k$, then $k$ is equal to:

• A. $122$
• B. $81$
• C. $119$
• D. $126$

Answer 1

Answer: (A)

$122$

$S_n=3+7+14+24+37+...+T_n$

$S_n=3+7+14+24+...+T_n$

$\therefore 0=3+4+7+10+13+...-T_n$

or $T_n=\frac{3+4+7+10+13+...}{rterm}$

$\therefore T_n=3+\left(\frac{n-1}{2}\right)[8+(n-2\left)3\right]$

$=3+\frac{(n-1)(3n+2)}{2}$

$=3+\frac{3n^2(3n+2)}{2}$

or $T_n=\frac{3}{2}{n}_2-\frac{n}{2}+2$

$\because S_n=\sum T_n$

$\therefore S_n=\frac{3}{2}\frac{n(n+1)(2n+1)}{6}-\frac{n(n+1)}{4}+2n$

$\therefore S_n=\frac{3}{2}\left[\frac{15\times 16\times 31}{6}\right]-\left(\frac{15\times 16}{4}\right)+(2\times 15)=1860-60+30$

$S_{15}=1830=15k$

$k=122$