23rd term of arithmetic sequence is17 and 17th term is 23 .what is the 40th term. how many terms of this sequence should to get a sum zero
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A17 =23. -(1)
A23=17. -(2)
Also,
A17=a+16d. (3)
Similarly A23=a+22d. (4)
By eliminating ( subtracting both )
a+16d=23
a+22d=17
-6d= 6
Hence d= -1
From equation (1)&(3)
a+16(-1)=23
a=23+16
a=39.
Now ,
A40= a+(40-1)-1
=39-39
=0
Hence 40 th term of this AP is “0”
Sn=n/2 [2a+(n-1)d]
0. = n/2[2(39)+(n-1)-1]
=n/2 [78-n+1]
=n/2[79-n]
=79n-n^2
n^2=79n
79=n^2/n
n=79
Sum of 79 terms of this AP is 0
A23=17. -(2)
Also,
A17=a+16d. (3)
Similarly A23=a+22d. (4)
By eliminating ( subtracting both )
a+16d=23
a+22d=17
-6d= 6
Hence d= -1
From equation (1)&(3)
a+16(-1)=23
a=23+16
a=39.
Now ,
A40= a+(40-1)-1
=39-39
=0
Hence 40 th term of this AP is “0”
Sn=n/2 [2a+(n-1)d]
0. = n/2[2(39)+(n-1)-1]
=n/2 [78-n+1]
=n/2[79-n]
=79n-n^2
n^2=79n
79=n^2/n
n=79
Sum of 79 terms of this AP is 0
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