Question

If the sum of the $12th$ and $22nd$ terms of an AP is $100$, then the sum of the first $33$ terms of an AP is
$A)1700$
$B)1650$
$C)3300$
$D)3400$

Answer 1

Hint: While talking about the A.P and G.P, we need to know about the concept of Arithmetic and Geometric progression.
An arithmetic progression can be represented by $a,(a + d),(a + 2d),(a + 3d),...$where $a$ is the first term and $d$ is a common difference.
A geometric progression can be given by $a,ar,a{r^2},....$ where $a$ is the first term and $r$ is a common ratio.
Here in this given question, we clearly see that the difference is $4$ and they give arithmetic progress.
Formula used:
The n-term of the AP is ${a_n} = a + (n - 1)d$
The sum of the n-terms of the A.P is ${S_n} = (\dfrac{n}{2})(2a + (n - 1)d)$

Complete step-by-step solution:
Since from the given that we have the sum of the $12th$ and $22nd$ terms of an AP are $100$,
First, the $12th$ term of the AP is denoted as ${a_{12}} = a + (12 - 1)d = a + 11d$
Similarly, the $22nd$ term of the AP is denoted as ${a_{22}} = a + (22 - 1)d = a + 21d$
Hence, we have the sum of the $12th$ and $22nd$ terms of an AP are $100$, thus we get ${a_{12}} + {a_{22}} = 100$ and substituting its values we have ${a_{12}} + {a_{22}} = 100 \Rightarrow a + 11d + a + 21d = 100$
Further solving we have, $2a + 32d = 100 \Rightarrow 2(a + 16d) = 100$
By division operation, we get $2(a + 16d) = 100 \Rightarrow a + 16d = 50$ which is the $17th$ term because which is in the form of ${a_{17}} = a + (17 - 1)d = a + 16d$
Since we need to find the sum of the n-term of $33$ terms.
Thus, the sum of the n-terms of the A.P is ${S_n} = (\dfrac{n}{2})(2a + (n - 1)d) \Rightarrow {S_{33}} = (\dfrac{{33}}{2})(2a + (33 - 1)d)$
Further solving we get ${S_{33}} = (\dfrac{{33}}{2})(2a + 32d) \Rightarrow 33(a + 16d)$ and since we know that $a + 16d = 50$
Hence, we get ${S_{33}} = 33(a + 16d) \Rightarrow 33 \times 50 = 1650$
Therefore, the option $B)1650$ is correct.

Note: For GP with the common ratio the formula to be calculated $GP = \dfrac{a}{{r - 1}},r \ne 1,r < 0$and $GP = \dfrac{a}{{1 - r}},r \ne 1,r > 0$
Harmonic progress is the reciprocal of the given arithmetic progression which is the form of $HP = \dfrac{1}{{[a + (n - 1)d]}}$where $a$ is the first term and $d$ is a common difference and n is the number of AP.