find the sum of the first 19 term of the AP whose 9th term is given by Tn = 7-4n
Answers
Answer:
-627
Step-by-step explanation:
Given : The nth term is given by t_n=7-4nt
n
=7−4n
To find : The sum of the first 19 terms of the AP ?
Solution :
The nth term is given by t_n=7-4nt
n
=7−4n
The first term, put n=1
t_1=7-4(1)=3t
1
=7−4(1)=3
The second term, put n=2
t_2=7-4(2)=-1t
2
=7−4(2)=−1
The common difference is d=t_2-t_1d=t
2
−t
1
d=-1-3=-4d=−1−3=−4
The sum of n number is given by,
S_n=\frac{n}{2}[2a+(n-1)d]S
n
=
2
n
[2a+(n−1)d]
The sum of 19 term, put n=19, a=3 and d=-4
S_{19}=\frac{19}{2}[2(3)+(19-1)(-4)]S
19
=
2
19
[2(3)+(19−1)(−4)]
S_{19}=\frac{19}{2}[6+(18)(-4)]S
19
=
2
19
[6+(18)(−4)]
S_{19}=\frac{19}{2}[6-72]S
19
=
2
19
[6−72]
S_{19}=\frac{19}{2}\times (-66)S
19
=
2
19
×(−66)
S_{19}=-627S
19
=−627
Therefore, the sum of the first 19 terms of the AP is -627.
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