Math, asked by arnavd52, 10 months ago


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

Answers

Answered by Anonymous
28

\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.

Answered by sanjeevk28012
0

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

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