Question

In an ap a=5 , d=3 ,an =50 find n and sn

Answer 1

Given:

• a (First term) = 5

• d (Common difference) = 3

• an (Final term) = 50

To Find:

• n and Sn

Solution:

We know that;

⇒ an = a + (n - 1)d

⇒ 50 = 5 + (n - 1)3

⇒ 50 - 5 = 3n - 3

⇒ 45 = 3n - 3

⇒ 45 + 3 = 3n

⇒ 48 = 3n

⇒ n = 48/3

⇒ n = 16

We know that;

$\Longrightarrow \sf S_n = \dfrac{n}{2} \Bigg(2a + (n-1)d\Bigg) \\ \\ \\ \\\Longrightarrow \sf S_{16} = \dfrac{16}{2} \Bigg(2(5) + (16-1)3\Bigg) \\ \\ \\ \\\Longrightarrow \sf S_{16} = 8 \Bigg(10 + (15)3\Bigg) \\ \\ \\ \\\Longrightarrow \sf S_{16} = 8 \bigg(10 + 45\bigg) \\ \\ \\ \\\Longrightarrow \sf S_{16} = 8 \bigg(55\bigg) \\ \\ \\ \\\Longrightarrow \sf S_{16} = 440$

∴ S$\sf _{n}$ = 440

Final Answers:

• n = 16

• Sn = 440

Answer 2

GIVEN :-

$\star \tt \:a \: = 5 \\ \star \tt \: d \: = 3 \\ \star \tt\: a_n = 50\:$

TO FIND :-

$\star$ $\tt{n}$

$\star$ $\tt{S_n}$

SOLUTION :-

As we know that,

$</p><p> \dagger \: { \boxed { \red{ \tt{a_n = a + ( n - 1)d}}}}$

[ Put the values ]

$\longrightarrow \: \tt \: 50 = 5 + ( n - 1)3 \\ \\ \longrightarrow \: \tt 50 =5 + 3n - 3 \\ \\ \longrightarrow \: \tt 50 - 5 + 3 = 3n \\ \\\longrightarrow \: \tt 48 = 3n \\ \\ \longrightarrow \: \tt n = \frac{ \cancel{48}}{ \cancel3} \\ \\ \longrightarrow \: \tt \huge \red { n \: =16 }$

Now,

$\dag \: { \boxed { \red{ \tt{S_n = \frac{n}{2} (2a + (n - 1)d)}}}}$

[ Put the values ]

$\longrightarrow \tt \: S_n = \frac{ \cancel{16}}{ \cancel2}[ (2 \times 5 + ( 16 - 1)]3 \\ \\\longrightarrow \tt \: S_n = 8 [10 + (15)3] \\ \\\longrightarrow \tt \: S_n = 8 [10 + 45] \\ \\ \longrightarrow \tt \: S_n = 8(55) \\\\ \longrightarrow \tt \: \huge \red{S_n = 440}$