Question

find the sum of first 12 terms of
A.P. 4,7,10……​

Answer 1

Given:-

AP:4,7,10,..

a=4

d=3

n=12

an=?

Answer:

an=a+(n-1)d

an=4+(12-1)3

an=33+4

an=37

so the 12 term of the given AP is 37

Answer 2

Step-by-step explanation:

Given:-

A.P : 4,7,10,...

To find:-

Find the sum of first 12 terms of an AP :4,7,10,...

Solution:-

Method-1:-

Given AP : 4,7,10,...

First term (a) = 4

Common difference (d)=7-4 = 3

We know that

The sum of first n terms of an AP is

Sn=(n/2)[2a+(n-1)d]

We have

a = 4

d = 3

n=12

On Substituting the values in the above formula

Sum of first 12 terms of the AP = S12

S12 = (12/2)[2(4)+(12-1)(3)]

=>S12 = (6)[8+11(3)]

=>S12 = 6(8+33)

=>S12 = 6(41)

=>S12 = 246

Method-2:-

We have a = 4

d = 3

n=12

We know that

The nth term of an AP = an = a+(n-1)d

an=4+(n-1)(3)

an=4+3n-3

an = 3n+1

a12=3(12)+1

a12=36+1

a12=37

Now Sum of first n terms=Sn =(n/2)[a+an]

=>S12 = (12/2)[4+37]

=>S12=(6)[41]

=>S12 = 246

Answer:-

The sum of the first 12 terms of the given AP is 246

Used formulae:-

• The sum of first n terms of an AP is
• Sn=(n/2)[2a+(n-1)d]
• Sum of first n terms=Sn =(n/2)[a+an]
• The nth term of an AP = an = a+(n-1)d
• a = First term
• d = Common difference
• n = Number of terms
• Sn = Sum of first n terms
• an = General or nth term