Question

Find the sum of the first 13 terms of an AP :-6, 0,6,12

Answer 1

Answer:

390

Step-by-step explanation:

Given AP: - 6 , 0 , 6 , 12 ...

 Here,

   First term = a = - 6

   Com. diff. = d = 12-6=6

From the properties of AP :

Sum of 1st n terms = (n/2)[2a+(n-1)d]

Therefore,

 ⇒ ( 13 / 2 )[ 2( - 6 ) + ( 13 - 1 )6 ]

 ⇒ ( 13 / 2 )[ - 12 + 12( 6 )

 ⇒ ( 13 / 2 )[ - 12 + 72 ]

 ⇒ ( 13 / 2 )[ 60 ]

 ⇒ 13( 30 )  

 ⇒ 390

Hence the sum of first 13 terms is 390

Answer 2

Step-by-step explanation:

Aloha !

$\text { This is Twilight Astro}$

We know that,..

Sum of 1 st n terms= ⁿ/2× 2a+(n-1)×d

Here,.

a is the first term in given numbers

d is different between given numbers

n=13

As we need 13 th term.

Now,.

=> n/2×2a+(n-1)×d

=>13/2×2(-6)+(13-1)×6

=>13/2×(-12)+(12)×6

=>13/2×(-12)+(72)

=>13/2×-12+72

=>13/2×60

=>13×30

=>390

Therefore,. 390 will be the sum of 13 th term for given A.P

Thank you

@ Twilight Astro ✌️☺️

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