Question

the sum of first five terms of the AP: 3,7,11,15....is ???​

Answer 1

a=3

d=4

n=5

Sum of n terms of AP = n/2[2a + (n – 1)d]

5/2[6+(4)4]

5/2 x 22

55

Answer 2

Answer:

55

Step-by-step explanation:

From the properties of AP :

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

 

Here,

   First term = a = 3

  Com. diff. = d = 7 - 3 = 4

Therefore.

⇒ Sum of first 5 terms

⇒ ( 5 / 2 )[ 2( 3 ) + ( 5 - 1 )4 ]

⇒ ( 5 / 2 ) [ 6 + ( 4 )4 ]

⇒ ( 5 / 2 ) [ 6 + 16 ]

⇒ ( 5 / 2 ) [ 22 ]

⇒ 5*11

⇒ 55

Hence the sum of the first 5 terms of the given AP is 55.