Question

in an ap:

(1)given a=2,d=3,an=50,find n and sn​

Answer 1

We have,

$\qquad \ \ \ a_n = 50 \\ \\ \implies \: a + (n - 1)d = 50 \\ \\ \implies 2 + (n - 1)3 = 50 \\ \\ \implies(n - 1)3 = 50 - 2 = 48 \\ \\ \implies(n - 1) = \frac{48}{3} = \frac{ \cancel{48} \: {}^{16} }{ \cancel{3} \: _1} \\ \\ \implies \: n - 1 = 16 \\ \\ \implies \: n = \bf 17$
Now,

$\quad \: S_n \\ \\ = \frac{n}{2} \left[ 2a + (n - 1)d \right] \\ \\ = \frac{17}{2} \left[ 2(2) + (17 - 1)(3) \right] \\ \\ = \frac{17}{2} \left[ 4 + 16(3) \right] \\ \\ = \frac{17}{2} (4 + 48) \\ \\ = \frac{17}{2} \times 52 = \frac{17}{ \cancel{2} \: _1} \times \cancel{52} \: {}^{26} \\ \\ = 17 \times 26 \\ \\ = \quad \bf442$
Hence,
$\underline{ \boxed{ \quad n = 17 \quad}}$
and,
$\underline{ \boxed{ \quad S_n = 442\quad }}$

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Additional Information
$\underline{\boxed{\bf{ \quad Arithmetic \quad Progression \quad}}}$
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In a sequence, if the difference between two consecutive terms is same then the sequence is known as an A.P.
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The first term is known as " a " .

The common difference is known as " d " .

The 'nth term' is given by : $\Large\sf\red{\underline{\fbox{\green{a_n = [a + (n - 1)d]}}}}$

The sum of 'n' terms is given by : $\Large\sf\red{\underline{\fbox{\green{S_n = \frac{n}{2} [2a + (n - 1)d]}}}}$
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