Question

find the sum of the first 12 terms of the arithmetic sequence whose general term is an=3n+5

Answer 1

Step-by-step explanation:

an=3n+5

a1=3x1+5=8

a2=3x2+5=11

a3=3x3+5=14

a12=3x12+5=41

Therefore it will be AP whose 12 terms are 8, 11, 14,........ 41

a=8, d=3, n=12, l=41

sn= n(a+1)/2=12(8+41)/2= 12x49/2=6x49=294

:the sum of 12 term are 294..

Answer 2

Sum of first 12 terms of sequence is 294

Given:

General term of a Arithmetic Sequence  is a$_{n}$=3n+5

To find:

Find the sum of the first 12 terms of the Arithmetic Sequence

Solution:

Given nth term of a sequence a$_{n}$=3n+5

⇒ first term of sequence (a₁) = 3(1)+5 = 3+5 = 8

⇒ Second term of sequence (a₂) = 3(2)+5 = 6+5 = 11

From 1st and 2nd terms

Common difference d = a₂-a₁ = 11 - 8 = 3  

As we know sum of n terms in AP S$_{n}$ = $\frac{n}{2} [ 2a+(n-1)d]$

⇒ Sum of first 12 terms S₁₂ = $\frac{12}{2} [ 2(8)+(12-1)3]$

= 6 [16+11(3)

= 6 [16+33]

= 294    

Sum of first 12 terms =  294

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