Question

the sum of first n term of AP is given by SN = 2n^2+ 3n and find the 16 term of AP

Answer 1

hello Frnd

Given as

$s_{n} = 2 {n}^{2} + 3n$
To find out we need to find first term and Common difference

Let n= 1

$s_{1} = 2 ({1}^{2} ) + 3(1) \\ = 5$
We got the first term a = 5

Now also put n=2

$s_{2} = 2( {2}^{2} ) + 3(2) \\ = 14$
Since it's the sum of 2 terms
That's why

2nd term = S2 - a
= 14 - 5
= 9

Now we have
a1 = 5 and a2 = 9

Now d = a2- a1
d = 9 - 5 = 4

Now 16th term

$a_{16} = a + 15d \\ = 5 + 15(4) \\ = 65$


hope it helps
jerri

Answer 2

hello,

akarsh here,

ur solution is,
.
.
.
s(n)=2n²+3n. (given)

first,
put the value of n =1

s(1)=2+3
=5
=a(first term)
.
.
.now,
put n=2......

s(2)=8+6
=14
.
to find a2(second term)
a2=s2-s1
=14-5
=9
.
.
d(common difference)=a2-a1
=9-5
=4

now,
a=5,d=4
.
.
.
therefore,
a(n)=a+(n-1)d

a(16)=5+(16-1)4

=60+5

=65

THEREFORE, 16TH TERM IS 65
.
.
.thank you