Question

Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.

Answer 1

Given that,

a₂ = 14

a₃ = 18

d = a₃ − a₂ 

= 18 − 14

= 4

a₂ = a + d

14 = a + 4

a = 10

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

S₅₁ = `51/2[2 xx 10 + (51 - 1)4]`

=`51/2[20 + (50)(4)]`

=`(51(220))/2`

= 51 × 110

= 5610

Answer 2

In the given problem, let us take the first term as a and the common difference as d

Here, we are given that,

a₂ = 14             ...(1)

a₃ = 18             ...(2)

Also we know

a_n = a + (n - 1)d

For the 2^nd term (n = 2)

a₂ = a + (2 – 1)d

14 = a + d    (Using 1)

a = 14 – d             ...(3)

Similarly for the 3rd term (n = 3)

a₃ = a + (3 - 1)d

18 = a + 2d        ...(Using 2)

a = 18 – 2d         ...(4)

Subtracting (3) from (4), we get,

a – a = (18 – 2d) – (14 – d)

0 = 18 – 2d – 14 + d

0 = 4 – d

d = 4

Now, to find a, we substitute the value of d in (4),

a = 14 – 4

a = 10

So for the given A.P d = 4 and a = 10

So, to find the sum of first 51 terms of this A.P., we use the following formula for the sum of n terms of an A.P.,

S_n = `"n"/2 [2"a" + ("n" - 1)"d"]`

Where; a = first term for the given A.P.

d = common difference of the given A.P.

n = number of terms

So, using the formula for n = 51, we get,

S₅₁ = `51/2 [2(10) + (51 - 1)(4)]`

= `51/2 [20 + (50)(4)]`

= `51/2 [20 + 200]`

= `51/2 [220]`

= 5610

Therefore, the sum of first 51 terms for the given A.P is S₅₁ = 5610.