Question

The third term of an A.P. is 7 and the seventh term exceeds three times the third term by 2. Find the first term, the common difference and the sum of first 20 terms.

Answer 1

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

Given,

a₃ = 7 …. (1) and,

a₇ = 3a₃ + 2   …. (2)

So, using (1) in (2), we get,

a₇ = 3(7) + 2 = 21 + 2 = 23  …. (3)

Also, we know that

a_n = a +(n – 1)d

So, the 3th term (for n = 3),

a₃ = a + (3 – 1)d

⟹ 7 = a + 2d   (Using 1)

⟹ a = 7 – 2d     …. (4)

Similarly, for the 7th term (n = 7),

a₇ = a + (7 – 1) d 24 = a + 6d = 23  (Using 3)

a = 23 – 6d  …. (5)

Subtracting (4) from (5), we get,

a – a = (23 – 6d) – (7 – 2d)

⟹ 0 = 23 – 6d – 7 + 2d

⟹ 0 = 16 – 4d

⟹ 4d = 16

⟹ d = 4

Now, to find a, we substitute the value of d in  (4), a =7 – 2(4)

⟹ a = 7 – 8

a = -1

Hence, for the A.P. a = -1 and d = 4

For finding the sum, we know that

S_n = \(\frac{n}{2}\)[2a + (n − 1)d] and n = 20 (given)

S₂₀  = \(\frac{20}{2}\)[2(−1) + (20 − 1)(4)]

= (10)[-2 + (19)(4)]

= (10)[-2 + 76]

= (10)[74]

= 740

Hence, the sum of first 20 terms for the given A.P. is 740