Question

find the sum of all even number between 1to150​

Answer 1

$\Huge{\underline{\underline{\blue{\mathfrak{Answer :}}}}}$

Given :

A.P :- 2, 4, 6, 8, 10 ........... 148

Note :- We doesn't considered 150 because we have to find the sum between 1 and 150 .

First term (a) = 2

Common Difference (d) = 2

Last term (L or An) = 148

$\rule{150}{2}$

To Find :

Sum of all terms of the A.P

$\rule{150}{2}$

Solution :

We know that,

${ \LARGE{ \leadsto{ \boxed{ \boxed{ \blue{ \sf{A_{n} \: = \: a \: + (n \: - \:1)d}}}}}}}$

(Putting Values)

148 = 2 + (n - 1)2

⟹148 - 2 = (n - 1)2

⟹ 146 = (n - 1)2

⟹ 146/2 = n - 1

⟹ 73 = n - 1

⟹ 73 + 1 = n

⟹ 74 = n

${ \LARGE{ \hookrightarrow{ \boxed{ \boxed{ \green{ \sf{n \: = \: 74}}}}}}}$

$\rule{200}{2}$

Now,

${ \LARGE{ \leadsto{ \boxed{ \boxed{ \blue{ \sf{S_{n} \: = \: \frac{n}{2} (a \: + \:L)}}}}}}}$

(Putting Values)

Sn = 74(2 + 148)/2

⟹ Sn = 74(150)/2

⟹ Sn = 11100/2

⟹ Sn = 5550

${\LARGE{\hookrightarrow{\boxed{\boxed{\green{\sf{S_{n} = 5550}}}}}}}$

Answer 2

Given ,

The AP is 2 , 4 , 6 , ...... , 148

Here ,

First term , a = 2

Common difference , d = 2

Last term , l = 148

We know that , the first nth term of an AP is given by

$\star \: \: \large\sf \fbox{A_{n} = a + (n - 1)d}$

Substitute the values , we obtain

$\sf \hookrightarrow 148 = 2 + (n - 1)2 \\ \\ \sf \hookrightarrow 146 = (n - 1)2 \\ \\ \sf \hookrightarrow 73 = n - 1 \\ \\ \sf \hookrightarrow n = 74$

Now , the sum of first n term of an AP is given by

$\star \: \: \large \sf \fbox{S_{n} = \frac{n}{2} (a + l)}$

Substitute the values , we obtain

$\sf \hookrightarrow S_{74} = \frac{74}{2} (2 + 148) \\ \\ \sf \hookrightarrow S_{74} = 37 \times 150 \\ \\ \sf \hookrightarrow S_{74} =5550$

Hence , the sum of first 74 even numbers is 5550