Question

19. The sum of first n terms of an AP is 3n² - 5, find the third term of the AP.​

Answer 1

$\huge{\underline{\tt{Given:-}}}$

Sum of first n terms of an AP = 3n²-5

$\huge{\underline{\tt{To\:Find:-}}}$

The third term = a3

$\huge{\underline{\tt{Solution:-}}}$

By using formula :-

${ \underline{ \boxed{ \bf{a _{n} = S _{n} - S _{n - 1}}}}} \: \: \: \: \: \: \: \: \: .....(1)$

${ \tt{S_{n} = (3n {}^{2} - 5)}} \: \: \: \: \: \: \: \: \: \: \: \: ....(2)\\ \\{ \tt {S _{n - 1}= 3(n- 1) {}^{2} - 5}}\\ \\{ \tt {S _{n - 1} = 3(n {}^{2} - 2n + 1) - 5 }}\\ \\ \: \: \: \: \: \: \: \: \: \: { = \tt{3n {}^{2} - 6n + 2}} \: \: \: \: \: \: \: \: \: ....(3)$

putting the equation (2) and (3) into (1)

we get,

${ \tt{a_{n} = (3n {}^{2} - 5) -(3n {}^{2} - 6n - 2)}}\\ \\{ \tt{ 3n {}^{2} - 5 - 3n {}^{2} + 6n + 2}}\\ \\{ \tt{ \cancel{ 3n {}^{2} }} - 5 - { \cancel{3n {}^{2} }} + 6n + 2}\\ \\ { \tt{ = 6n - 3}}\\ \\{ \tt{ n = 3 \: \: (3rd \:( term \: given))}}\\ \\{ \tt{ putting \: the \: value \: we \: get}}\\ \\ { \tt{= 6(3) - 3}} \\\\ { \tt{ = 18 - 3}}\\ \\ = { \underline{ \overline{ \bf{ \mid{15} \mid}}}}$

•°• 15 is the 3rd term of the given AP.

Answer 2

Answer:

The third term of AP is 5 .

Step-by-step explanation:

Given as :

The sum of first n terms of an AP is 3 n² - 5

For Arithmetic progression

Sum of n term = $S_n$ = $\dfrac{n}{2}$ [ 2 a + ( n - 1 ) d ]

$S_n$  = 3 n² - 5

$S_n-_1$ = 3 (n – 1)² + 5

Or, $S_n-_1$ = 3 × [ n² + 1 - 2 n ] + 5

Or, $S_n-_1$ = 3 n²  -  6 n + 3 + 5

∴ $S_n-_1$ = 3 n² - 6 n + 8

Again

$a_n$ term  = $S_n$  - $S_n-_1$

i.e  $a_n$ = ( 3 n² - 5) - (  3 n² - 6 n + 8 )

Or, $a_n$ = ( 3 n² - 3 n² ) + 6 n - 8 - 5

Or, $a_n$  = 0 + 6 n - 13

∴  $a_n$ = 6 n - 13

So, The nth term = 6 n - 13

Now, for n = 1

$a_1$  = 6 × 1 - 13 = - 7

similarly

$a_3$   = 6 × 3 - 13 = 5

So, The third term = $a_3$  = 5

Hence, The third term of AP is 5 . Answer