Question

In an AP the sum of first n terms is 3n²/2+5n/2 find its 25 th term

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Answer 1

☯ It is given that, Sum of first n terms of an AP is $\sf \dfrac{3n^2}{2} + \dfrac{5n}{2}$

☯ We have to find, $\sf 25^{th}$ term of AP.

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$\dag\;{\underline{\frak{We\;know\;that\;:}}}$

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$\star\;{\boxed{\sf{\pink{n^{th}\;term = S_n - S_{(n - 1)}}}}}$

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Therefore,

$:\implies\sf a_{25} = S_{25} - S_{(25 - 1)}\;\;\;\;\;\;\;\;\bigg\lgroup\bf eq.\;(1) \bigg\rgroup$

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$\dag\;{\underline{\frak{We\;have\;:}}}$

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$:\implies\sf S_n = \dfrac{3n^2}{2} + \dfrac{5n}{2}$

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$\dag\;{\underline{\frak{Putting\;n = 25\;:}}}$

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$\;\;:\implies\sf S_{25} = \dfrac{3(25)^2}{2} + \dfrac{5(25)}{2}$

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$:\implies\sf S_{25} = \dfrac{3 \times 625}{2} + \dfrac{5 \times 25}{2}$

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$\;\;\;\;\;\::\implies\sf S_{25} = \dfrac{1875}{2} + \dfrac{125}{2}$

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$\;\;\;\;\;\;:\implies\sf S_{25} = \dfrac{1875 + 125}{2}$

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$\;\;\;\;\;\;\;\;\;:\implies\sf S_{25} = \dfrac{2000}{2}$

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$\;\;\;\;\;\;\;\;\;:\implies\sf S_{25} = \cancel{ \dfrac{2000}{2}}$

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$\;\;\;\;\;\;\;:\implies{\underline{\boxed{\sf{\pink{S_{25} = 1000}}}}}\;\bigstar$

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${\underline{\sf{\bigstar\;Again\;putting\;n = 24}}}$

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$\;\;\;:\implies\sf S_{24} = \dfrac{3(24)^2}{2} + \dfrac{5(24)}{2}$

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$:\implies\sf S_{24} = \dfrac{3 \times 576}{2} + \dfrac{5 \times 24}{2}$

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$\;\;\;\;\;\;:\implies\sf S_{24} = \dfrac{1728}{2} + \dfrac{120}{2}$

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$\;\;\;\;\;\;:\implies\sf S_{24} = \dfrac{1728 + 120}{2}$

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$\;\;\;\;\;\;\;\;\;\;:\implies\sf S_{24} = \dfrac{1848}{2}$

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$\;\;\;\;\;\;\;\;\;:\implies\sf S_{24} = \cancel{ \dfrac{1848}{2}}$

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$\;\;\;\;\;\;\;\;:\implies{\underline{\boxed{\sf{\pink{S_{24} = 924}}}}}\;\bigstar$

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$\dag\;{\underline{\frak{Now,\;We\;have\;:}}}$

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• $\sf S_{25} = 1000$

• $\sf S_{24} = 924$

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$\dag\;{\underline{\frak{Putting\; values\;in\;eq(1)\;:}}}$

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$:\implies\sf a_{25} = S_{25} - S_{(25 - 1)}$

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$:\implies\sf a_{25} = 1000 - 924$

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$\;\;\;\;\;\;:\implies{\underline{\boxed{\sf{\purple{a_{25} = 76}}}}}\;\bigstar$

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$\therefore\;\sf a_{25}$ term of AP is 76.

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$\boxed{\bf{\mid{\overline{\underline{\bigstar\: Formulae\:related\:to\:AP :}}}}\mid}$

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• $\sf n^{th}\;term\;of\;AP = a_n = a + (n - 1)d$

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• $\sf Sum\;of\;n\;terms\;of\:AP = S_n = \frac{n}{2} \bigg\lgroup\sf 2a + (n - 1)d \bigg\rgroup$

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• $\sf Sum\;of\;AP\;having\;last\;term\;as\;'l' = S_n = \frac{n}{2}(a + l)$

Answer 2

Solution :

It is given that sum of n terms of A.P. is $\sf \dfrac{3n^2}{2} + \dfrac{5n}{2}$

Let the sum of n terms is given by $\sf S_n$

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So,

$:\implies \sf S_n = \Bigg \lgroup\dfrac{3n^2}{2} + \dfrac{5n}{2} \Bigg \rgroup$

Now, put n = 1 :

$:\implies \sf S_1 = \Bigg \lgroup\dfrac{3(1)^2}{2} + \dfrac{5(1)}{2} \Bigg \rgroup$

$:\implies \sf S_1 = \Bigg \lgroup\dfrac{3}{2} + \dfrac{5}{2} \Bigg \rgroup$

$:\implies \sf S_1 = 4$

We know, $\sf S_1 = a_1$ as $\sf S_1$ is a sum of first term only.

Therefore, the first term is a = 4.

Now, Put n = 2 :

$:\implies \sf S_2 = \Bigg \lgroup\dfrac{3(2)^2}{2} + \dfrac{5(2)}{2} \Bigg \rgroup$

$:\implies \sf S_2 = \Bigg \lgroup\dfrac{3(4)}{2} + \dfrac{5(2)}{2} \Bigg \rgroup$

$:\implies \sf S_2 = \Bigg \lgroup\dfrac{12}{2} + \dfrac{10}{2} \Bigg \rgroup$

$:\implies \sf S_2 =6 + 5$

$:\implies \sf S_2 = 11$

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As we know that,

$\bigstar\:\: \sf S_2 = a_1 + a_2\:\:\bigstar$

$\dashrightarrow\:\: \sf 11 = 4 + a_2$

$\dashrightarrow\:\: \sf a_2 = 11 - 4$

$\dashrightarrow\:\: \sf a_2 = 7$

As,

$\sf a_1 = 4$ and $\sf a_2 = 7$

So, common difference (d) will be :

$\bigstar \: \: \sf d = a_2 - a_1$

$: \implies \sf d = 7 - 4$

$: \implies \sf d = 3$

Thus,

$\bigstar\:\: \sf a_n = a + (n - 1) d\:\:\bigstar$

$\leadsto \sf a_{25} =4 + (25 - 1) \times 3$

$\leadsto \sf a_{25} =4 + 24\times 3$

$\leadsto \sf a_{25} =4 + 72$

$\leadsto \sf a_{25} =76$

$\therefore\:\underline{\textsf{The 25th term of the AP is \textbf{76}}}.$

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$\boxed{\underline{\underline{\bigstar \: \bf\:Extra\:Brainly\:knowlegde\:\bigstar}}}$

Que : What is Arithmetic Progression ?

Ans : It is a mathematical sequence in which difference is between two consecutive terms is always constant is called as$\sf \underline{\red {Arithmetic \: Progression.}}$

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$\boxed{\underline{\underline{\bigstar\:\bf\:Some \: important \: Formulae \: related \: to \: it \:\bigstar}}} \\ \\ </p><p></p><p>$

$\sf (i) \: The\: nth\: term\: of \:an \:AP:\:\:\:\: \red{a_n = a + (n - 1) d}$

$\sf(ii) \: Sum \: of \: n \: terms \: in \: AP :\:\:\:\: \purple{S = \dfrac{n}{2}\Bigg\{ 2a + (n-1) d\Bigg\}}$

$\sf (iii ) \: Sum \: of \: all \: terms \: in \: a \: finite \: AP \: with \: the \: last \: term \: as \: 'l': \:\:\:\: \pink{\dfrac{n}{2} (a + l)}$

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