Answer:
1632
Step-by-step explanation:
given
a= 82
d= -2
an= 20
n=?
sn=?
an= a+(n-1)d
20= 82+(n-1)-2
20= 82+ (-2n)+2
20-84= -2n
-64/-2= n
n= 32
sn= n/2 (2a+(n-1)d
sn= 32/2 (2*82 + (32-1) -2
sn= 16 ( 164 + 31*(-2)
sn= 16 (164-62)
sn= 16*102
sn= 1632
The given series in srithmetic progression:
82 + 80 + 78 +.............+ 20
To find, the sum of the series,
Here, first term(a) = 82, common difference(d) = 80 -82 = - 2 and
last term() = 20
We know that,
⇒
⇒ 2n = 64
⇒ n = 32
The sum of the series,
= 1632
Hence, the sum of the arithmetic progression is "1632".
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Answers & Comments
Answer:
1632
Step-by-step explanation:
given
a= 82
d= -2
an= 20
n=?
sn=?
an= a+(n-1)d
20= 82+(n-1)-2
20= 82+ (-2n)+2
20-84= -2n
-64/-2= n
n= 32
sn= n/2 (2a+(n-1)d
sn= 32/2 (2*82 + (32-1) -2
sn= 16 ( 164 + 31*(-2)
sn= 16 (164-62)
sn= 16*102
sn= 1632
The sum of the arithmetic progression is "1632".
Step-by-step explanation:
The given series in srithmetic progression:
82 + 80 + 78 +.............+ 20
To find, the sum of the series,
Here, first term(a) = 82, common difference(d) = 80 -82 = - 2 and
last term(
) = 20
We know that,
⇒![20=82+(n-1)(-2) 20=82+(n-1)(-2)](https://tex.z-dn.net/?f=20%3D82%2B%28n-1%29%28-2%29)
⇒![20=82-2n+2=84-2n 20=82-2n+2=84-2n](https://tex.z-dn.net/?f=20%3D82-2n%2B2%3D84-2n)
⇒ 2n = 64
⇒ n = 32
The sum of the series,![S_{n}=\dfrac{n}{2} (a+a_{n}) S_{n}=\dfrac{n}{2} (a+a_{n})](https://tex.z-dn.net/?f=S_%7Bn%7D%3D%5Cdfrac%7Bn%7D%7B2%7D%20%28a%2Ba_%7Bn%7D%29)
= 1632
Hence, the sum of the arithmetic progression is "1632".