The third term of an A.P. is 7 and the seventh term exceeds three time

1.	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» 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

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»

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