The sum of first m terms of an ap is4m^2-m if its nth term is 107 find the value of n also find its 21st term of the ap
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Given ,Sum of m terms =4m²-m
Now Sum of 1 term =4(1)²-1=3
Sum of terms =4(2²)-2=14
First term =3
Second term =14-3=11
Common difference =11-3=8
Now ,We know that nth term = a1 + (n – 1) d.
Given nth term =107
107=3+(n-1)8
107-3=8n-8
104+8=8n
112/8=n
14=n
Therefore,107 is the 14 th term .
21 st Term of A.P =3+(21-1)8=3+20*8=163
Therefore 21 st Term =163 .
Now Sum of 1 term =4(1)²-1=3
Sum of terms =4(2²)-2=14
First term =3
Second term =14-3=11
Common difference =11-3=8
Now ,We know that nth term = a1 + (n – 1) d.
Given nth term =107
107=3+(n-1)8
107-3=8n-8
104+8=8n
112/8=n
14=n
Therefore,107 is the 14 th term .
21 st Term of A.P =3+(21-1)8=3+20*8=163
Therefore 21 st Term =163 .
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