The sum of first n terms of an ap is 3n^2 + 6n. find the nth term of this ap
Solution:- Given that An(nth term of an AP) =3n^2+6n
We know that
An=Sn - S(n-1)
=3n^2+6n -{3(n-1)^2+6(n-1)}
=3n^2+6n -{3(n^2 -2n +1)+6(n-1)}
=3n^2+6n -{3n^2 - 6n +3+ 6n -6}
=3n^2+6n -{3n^2 +3-6}
=3n^2+6n -{3n^2 -3}
=3n^2+6n -3n^2 +3
=6n+3 Ans....
Question 11 .If the sum of the first n terms of an AP is 4n − n^2 , what is the first term (that is a1)? What is the sum of first two terms? What is the second term? Similarly find the 3 rd , the10 th and the n th terms.
solution
Sum of the nth terms of an AP(Sn)=4n-n^2
An(the n th terms of the Ap)
an=Sn - Sn-1(proof see Video-click here)
= 4n-n^2 -{4(n-1)-(n-1)^2}
= 4n-n^2-{4n-4-(n^2-2n+1)}
=4n-n^2-{4n-4-n^2+2n-1)}
Sum of the n terms of an A.P. is 5n^2 - 3n. Find the terms of the A.P. and also find the 16th term.
Sum of the nth terms of an AP(Sn)=5n^2 - 3n
An(the n th terms of the Ap)
an=Sn - Sn-1
= 5n^2 - 3n-{5(n-1)^2 - 3(n-1)}
= 5n^2 - 3n-{5(n^2-2n+1) - 3(n-1)}
=5n^2 - 3n-{5n^2-10n+5 - 3n+3}
=5n^2 - 3n - 5n^2+10n-5 + 3n-3}
an=10n-8
a1=10×1-8
a1=10-8
=2
now common difference(d)=an-an-1
d=10n-8 -{10(n-1)-8}
d=10n-8 -{10n-10-8}
d= 10n -8- 10n+10 +8
d=10
term of the AP
a1=a= 2
a2=a+d= 2+10=12
a3=a2+d=12+10=22
now,sequence of AP is
2,12,22.........
16 th term of the given AP=
a16=10×16-8
a16=160-8 =152
