Question

In an arithmetic progression an = 10-3n find S40?
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Answer 1

Answer:

$S_{40}=-2060$

Step-by-step explanation:

Formula used:

The sum of n terms of an A.P a, a+d, a+2d,.... is

$S_n=\frac{n}{2}[a+l]$

Given:

$a_n=10-3n$

Then,

$a=a_1=10-3=7$

$l=a_{40}=10-3(40)=10-120=-110$

Now,

$S_n=\frac{n}{2}[a+l]$

$S_{40}=\frac{40}{2}[7-110]$

$S_{40}=20[-103]$

$S_{40}=-2060$

Answer 2

The sum of 40 term of an A.P is -2060.

Step-by-step explanation:

Given : In an arithmetic progression $a_n = 10-3n$.

To find : The value of $S_{40}$ ?

Solution :

The nth term of A.P is $a_n = 10-3n$

Put n=1 for first term,

$a_{1}= 10-3(1)=7$

Put n=2 for second term,

$a_{2}= 10-3(2)=4$

The common difference is $d=a_2-a_1$

$d=4-7$

$d=-3$

The sum of n term of an A.P is

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

The sum of 40 term of an A.P is

$S_{40}=\frac{40}{2}[2(7)+(40-1)(-3)]$

$S_{40}=20[14-117]$

$S_{40}=20[-103]$

$S_{40}=-2060$

The sum of 40 term of an A.P is -2060.

#Learn more

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