Question

find the sum of first 24 terms of the list no whose nth term is given by an = 3-2n​

Answer 1

$\large{\orange{\bf{\underline{\underline{Answer:}}}}}$

$\dag \large{\boxed{\mathfrak{Arithmetic\;Progession}}}$

Here, as we already have nth term that is 3 - 2n . So, from this we can find the starting terms of AP :

➺ Put 1, 2, 3 in the given equation :

( 1st term )

⇝${\sf{{a}_{1}}}$ = 3 - 2n

⇝${\sf{{a}_{1}}}$ = 3 - 2( 1 )

⇝${\sf{{a}_{1}}}$ = 3 - 2

⇝${\sf{{a}_{1}}}$ = 1

( 2nd term )

⇝${\sf{{a}_{2}}}$ = 3 - 2n

⇝${\sf{{a}_{2}}}$ = 3 - 2( 2 )

⇝${\sf{{a}_{2}}}$ = 3 - 4

⇝${\sf{{a}_{2}}}$ = - 1

( 3rd term )

⇝${\sf{{a}_{3}}}$ = 3 - 2n

⇝${\sf{{a}_{3}}}$ = 3 - 2( 3 )

⇝${\sf{{a}_{3}}}$ = 3 - 6

⇝${\sf{{a}_{3}}}$ = - 3

Thus , the terms of the AP are 1 , - 1 , - 3.

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From above , we can say that the first term if the AP is 1 and the common difference is - 2

★ Formula for Arithmetic Progession :

$\large{\boxed{\bf{{S}_{n} = \dfrac{ n }{2} [ 2a + ( n - 1 ) d ] }}}$

So, put the value of a = 1 and d = - 2 in the formula :-

$\hookrightarrow$ ${\bf{{S}_{n} = \dfrac{ n }{2} [ 2a + ( n - 1 ) d ] }}$

$\hookrightarrow$ ${\bf{{S}_{24} = \dfrac{ 24 }{2} [ 2(1) + ( 24 - 1 ) - 2 ] }}$

$\hookrightarrow$ ${\bf{{S}_{24} = 12 [ 2 + ( 23 ) - 2 ] }}$

$\hookrightarrow$ ${\bf{{S}_{24} = 12 [ 2 + ( - 46 ) ] }}$

$\hookrightarrow$ ${\bf{{S}_{24} = 12 [ 2 - 46 ) ] }}$

$\hookrightarrow$ ${\bf{{S}_{24} = 12 × - 44 }}$

$\hookrightarrow$ ${\bf{{S}_{24} = -528 }}$

•°• The sum of first 24 terms of the list number is -528.

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