Question

hlo frnds
$\: the \: sum \: of \: frst \: n \: \: term \: of \: an \: ap \: is \: (3 {n}^{2} + 4 n). \: find \: \: its \: n \: th \: term \: and \: its \: ap$
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Answer 1

Solution :-

$<b><i>$

Sn = 3n² + 4n

Put, n = 1 ,2,3,4..........

#N=1

$\huge{\bf{\implies}}$ 3n² + 4n

$\huge{\bf{\implies}}$ 3(1)²+4

$\huge{\bf{\implies}}$ 7

#put n = 2

$\huge{\bf{\implies}}$ 3(2)²+4(2)

$\huge{\bf{\implies}}$ =20

#put n = 3

$\huge{\bf{\implies}}$ 3(3)²+4

$\huge{\bf{\implies}}$ = 31

S1 = 7

S2 = 20

S3 = 31

$\bf{\implies}$D = 6

$\bf{\implies}$A = 7

$\huge{\bf{\implies}}$ An = A + (n-1)D

= 7 + (n-1)6

An= 6n +1

putting n=1,2,3,4....
you will get its terms

A.P ----> 7,13,19......

Answer 2

Step-by-step explanation:

we know,

Sn = n/2{ 2a + ( n -1)d}

now, given

Sn = 3n² + 4n

=n/2{ 6n + 8}

=n/2{ 2.7 + ( n -1)6}

now you see this is same as ,above formula in which a = 7

d = 6

Tn = 7 + 6( n -1)

=6n +1

put T1 = 6×1 +1 =7

T2 = 6×2 + 1 = 13

T3 =6×3 +1 =19

13 -7 = 19 -13 =6

so, this is an AP