Question

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

Answer 1

Let’s take the first term as a and the common difference as d.

Given that,

a₂ = 14 and a₃ = 18

And, we know that a_n = a + (n – 1)d

So,

a₂ = a + (2 – 1)d

⟹ 14 = a + d

⟹ a = 14 – d …. (i)

Similarly,

a₃ = a + (3 – 1)d

⟹ 18 = a + 2d

⟹ a = 18 – 2d …. (ii)

Subtracting (i) from (ii), we have

a – a = (18 – 2d) – (14 – d)

0 = 18 – 2d – 14 + d

0 = 4 – d

d = 4

Putting d in (i), to find a

a = 14 – 4

a = 10

Thus, for the A.P. d = 4 and a = 10

Now, to find sum of terms

S_n = \(\frac{n}{2}\)(2a + (n − 1)d)

Where,

a = the first term of the A.P.

d = common difference of the A.P.

n = number of terms So, using the formula for

n = 51,

⟹ S₅₁  = \(\frac{51}{2}\)[2(10) + (51 – 1)(4)]

= \(\frac{51}{2}\)[20 + (40)4]

= \(\frac{51}{2}\)[220]

= 51(110)

= 5610

Hence, the sum of the first 51 terms of the given A.P. is 5610