Question

Answer this qn from Arithmetic progression chapter​

Answer 1

Answer:

Sum of first thirty terms of the AP whose general term is given by $a_{n} =7+2n$ is $S_{30} =1140$ .

Step-by-step explanation:

An Arithmetic Progression is a sequence of numbers which either increases or decreases with a constant integer. That constant integer is called the common difference of the progression and is denoted by d.

The nth term is the general term of the progression which is represented by $a_{n}$. If the general term is given, then we can find the progression.

So in this question the general term is given by

$a_{n} =7+2n$ .

So the first term is    $a_{1} =7+2(1)=9$

second term             $a_{2} =7+2(2)=11$

third term                  $a_{3} =7+2(3)=13$    etc.

so the progression is    $9,11,13,....$

We want to calculate the sum of first $30$ terms of the progression.

We have the formula for finding the sum of first n terms of the progression

$S_{n} =\frac{n}{2} [2a+(n-1)d]$

here  $n=30,\ a=9,\ d=11-9=2,\ n-1=29$

substituting all these values we get,

$S_{30} =\frac{30}{2} [2*9+29*2]$

     $=15[18+58]\\\\=15*76\\\\=1140$

Therefore the sum of first thirty terms of the AP is $1140$ .

Hence the answer

thank you

Answer 2

Answer:

$an = 7 + 2n$

so,

first number

$a1 = 7 + 2 = 9$

last or 30th term,

$a30 = 7 + 2 \times 30 = 67$

now the sum of first 30 terms is,

$s = \frac{n}{2} (a1 + a30) = \frac{30}{2}(9 + 67) \\ = 15 \times 76 = 1140$