Question

Match the arithmetic progressions with their sums.

• A. 10100
• B. 40703
• C. 10000

Answer 1

We know that sum to $n$ terms of an AP is given by
$S_n=\frac{n}{2}(a+l),$ where $a$ and $l$ are the first and the last terms respectively, and $n$ is the number of terms in the AP.

In the AP $2+4+...+200,$
$a=2,d=2$ and $l=200.$
The $n^{\text{th}}$ term of the AP is given by
$a_n=a+(n-1)d.$
$\therefore 200=2+(n-1)2$
$198=(n-1)2$
$⟹99=(n-1)$
$⟹n=100$
$\therefore S_n=\frac{100}{2}(2+200)$
$S_n=10100$

Similarly, in the AP 3 + 11 + ...+803,
$a=3,d=8$ and $l=803.$
Using $a_n=a+(n-1)d$ to calculate $n,$ we have
$803=3+(n-1)8$
$800=(n-1)8$
$⟹n=101$
$⟹S_n=\frac{101}{2}(3+803)$
$\therefore S_n=40703$

Lastly, in the AP 1 + 3 +...+199, we have
$a=1,d=2$ and $l=199.$
Using $a_n=a+(n-1)d,$
$199=1+(n-1)2.$
$198=(n-1)2$
$99=(n-1)⟹n=100$
$⟹S_n=\frac{100}{2}(1+199)$
$\therefore S_n=10000$