Question

Find the sum of 46 + 42 + 38 + ... + (-446) + (-450)46+42+38+...+(−446)+(−450)

Answer 1

sum of sequence Find the sum of 46 + 42 + 38 + ... + (-446) + (-450) is -25,250

Step-by-step explanation:

We need to find sum of sequence  : 46 + 42 + 38 + ... + (-446) + (-450)

Given sequence is an AP with following parameters as :

$a=46\\d=42-46=-4$

So , Let's calculate how many terms are there as :

⇒ $a_n=a +(n-1)d$

⇒ $-450=46 +(n-1)(-4)$

⇒ $-496=(n-1)(-4)$

⇒ $\frac{-496}{-4}=n-1$

⇒ $124=n-1$

⇒ $n=125$

Sum of an AP is :

⇒ $S_n = \frac{n}{2}(2a+(n-1)d)$

⇒ $S_1_2_5 = \frac{125}{2}(2(46)+(125-1)(-4))$

⇒ $S_1_2_5 = \frac{125}{2}(-404)$

⇒ $S_1_2_5 =-25,250$

Therefore , sum of sequence Find the sum of 46 + 42 + 38 + ... + (-446) + (-450) is -25,250

Answer 2

The sum of the given sequence is -25500.

Step-by-step explanation:

The given Arithmetic sequence is 46 + 42 +38... +(-446) +(-450).

• The first term of the sequence = 46
• The last term of the sequence = -450
• The common difference ⇒ 42 - 46 = - 4

To find the number of terms in the sequence :

The formula used is $n = (\frac{a_{n}-a_{1}} {d})+1$

where,

• n is the number of terms.
• $a_{n}$ is the late term which is -450.
• $a_{1}$ is the first term which is 46.
• d is the common difference which is 4.  

Therefore, $n =(\frac{-450-46}{-4}) +1$

⇒ $n = (\frac{-496}{-4}) + 1$

⇒ $n = 124 + 1$

⇒ $n =125$

∴ The number of terms, n = 125.

To find the sum of the arithmetic progression :

The formula used is $S = \frac{n}{2}(a_{1} + a_{n} )$

where,

• S is the sum of the sequence.
• $a_{1}$ is the first term which is 46.
• $a_{n}$ is the late term which is -450.

Therefore, $S = \frac{125}{2}(46+ (-450))$

⇒  $S = \frac{125}{2}(-404)$

⇒ $S = 125 \times -202$

⇒ $S = -25500$

∴ The sum of the given sequence is -25500.