If the sum of the first n terms of an A.P is 4n^2 then the difference between the 3rd and 10th term is
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S
n
=4n−n
2
S
1
=4−1=3
But Sum of First Term is the First Term
So, T
1
=a=3
and, Sum of first two terms is S
2
=8−4=4
Second Term T
2
=S
2
−S
1
=4−3=1
Now, common difference =d=T
2
−T
1
=1−3=−2
Now, T
n
=a+(n−1)d=3+(n−1)(−2)=3+2−2n=5−2n
Hence, T
2
=1
T
3
=−1
T
10
=5−2∗10=5−20=−15
T
n
=5−2n
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