Question

Add : $7+10\frac{1}{2}+14+.......+84$

Answer 1

Add $7+101/2+\mathrm{14.......84}$

The above series forms an Arithmetic progression (A.P) with first term $a=7$ & the common difference $d=3.5$

we know that the $n^{th}$ term of A.P is given by

$a+(n-1)d$

here $a+(n-1)d=84$

$7+(n-1)3.5=84$

$7+(n-1)3.5=84$

$7+(n-1)3.5-7=84-7$ [ subtract $7$ from sides]

$(n-1)3.5=77$

divide both sides by $3.5$

$(n-1)=77/3.5$

$(n-1)=\frac{77\times 10}{35}$

$(n-1)=22$

Add $1$ on both sides

$n=23$

$\Rightarrow 84$ is the ${23}^{rd}$ term

Sum of $n$ terms of A.P is given by

$S_n=n/2p2a+(n-1)d]$

$n=23,a=7,d=3.5$

$\Rightarrow S_n=\frac{23}{2}\left[2\right(7)+(23-1\left)3.5\right]$

$S_n=\frac{23}{2}[14+22(3.5\left)\right]$

$S_n=\frac{23}{2}[14+77]$

$S_n=\frac{23}{2}\left[91\right]$

$S_n=\frac{23}{2}\times 91$

$S_n=1046.5$