Question

Find the sum of the 51 terms of an ap whose second and third terms of 14 and 15 respectively

Answer 1

In Arithmetic Progressions,

a = first term

d = common difference

a₂ = second term = a + d

aₓ = x th term

============================

Now,

Given that 2nd and 3rd term of the AP are 14 and 15.

∴ 2nd term = a + d = 14

                            a = 14 - d      ...( i )

∴ 3rd term = a + 2d = 15  

                             a = 15 - 2d  ...( ii )

Comparing the values of a from ( i ) & ( ii )

⇒ 14 - d = 15 - 2d

⇒ 2d - d = 15 - 14

⇒ d = 1

∴ Common difference of the given AP is 1.

Substituting the value of d in ( 1 )

⇒ a = 14 - d

⇒ a = 14 - 1

⇒ a = 13

∴ First term of the AP is 13.

$\maths{\text{From the properties of Arithmetic Progressions}}, \\ \mathsf{S_{n} = \dfrac{n}{2}\bigg[ 2a + ( n - 1 )d \bigg] }$

Where n is the number of terms.

∴ S₅₁ = $\dfrac{51}{2}\bigg[ 2(13) + (51-1)1 \bigg]$

⇒ S₅₁ = $\dfrac{51}{2} \bigg[ 26 + 50 \bigg]$

⇒ S₅₁ = $\dfrac{51}{2} \times 76$

⇒ S₅₁ = 51 x 38

⇒ S₅₁ = 1938

Therefore the sum of 51 terms of the given AP is 1938.