Question

find the 50th term and sum of first 50 terms of A.P 11,16,21,26............​

Answer 1

Given: An arithmetic progression.

To find: The 50th term and the sum of the first 50 terms.

Answer:

AP - 11, 16, 21, 26, ...

From this progression,

• The first term (a) = 11.
• The common difference (d) = 16 - 11 = 5.

Let's find the 50th term.

$\sf a_{50}\ =\ a\ +\ 49\ \times\ d\\\\\\a_{50}\ =\ 11\ +\ 49\ \times\ 5\\\\\\a_{50}\ =\ 11\ +\ 245\\\\\\a_{50}\ =\ 256$

Therefore, the 50th term is 256.

Now, let's find the sum of the first 50 terms.

$\sf S_{50}\ =\ \dfrac{n}{2}\ \times\ \bigg[2a\ +\ (n\ -\ 1)d\bigg]\\\\\\S_{50}\ =\ \dfrac{50}{2}\ \times\ \bigg[2\ \times\ 11\ +\ (50\ -\ 1)(5)\bigg]\\\\\\S_{50}\ =\ 25\ \times\ \bigg[22\ +\ 245\bigg]\\\\\\S_{50}\ =\ 25\ \times\ 267\\\\\\S_{50}\ =\ 6675$

Therefore, the sum of the first 50 terms is 6675.

Answer 2

AnsweR:

Given:

• a = 11
• d = 16-11 = 5
• AP = 11,16,21,26....

To Find:

• $\tt{a_n = ?}$ (if n is 50)
• $\tt{S_n = ?}$ (if n is 50)

Solution:

• $\tt{a_n = ?}$ (if n is 50)

Formula: $\tt{a_n = a + (n - 1)d}$

➜$\tt{a_{50} = 11 + (50 - 1)5}$

➜$\tt{a_{50} = 11 + (49)5}$

➜$\tt{a_{50} = 11 + 245}$

➜$\tt{\green{a_{50} = 256}}$

• $\tt{S_n = ?}$ (if n is 50)

Formula: $\tt{S_n = \dfrac{n}{2} \bigg( 2a + (n - 1)d \bigg)}$

➜$\tt{S_{50} = \dfrac{50}{2} \bigg( 2(11) + (50 - 1)5 \bigg)}$

➜$\tt{S_{50} = \dfrac{50}{2} \bigg( 22 + (49)5 \bigg)}$

➜$\tt{S_{50} = \dfrac{50}{2} \bigg( 22 + 245 \bigg)}$

➜$\tt{S_{50} = \dfrac{50}{2} ( 267 )}$

➜$\tt{S_{50} = \dfrac{50}{2} ( 267 )}$

➜$\tt{S_{50} = 25 \times ( 267 )}$

➜$\tt{\green{S_{50} = 6,675}}$

$\rule{200}2$