Find the sum given below:

7 + 10 + 13 + ... + 46

OR

If the 9th term of an A.P. is zero, then show that its 29th term is double of its 19th term.

Given: the first term, a = 7

last term, l = 46

and common difference, d = a₂ – a₁ = 10 – 7 = 3

To find: Sum of 7 + 10 + 13 + ... + 46

Proof: We know that,

Firstly, we find n

a_n = a + (n – 1)d

Here, a_n = l = 46

⇒ 46 = 7 + (n – 1)×3

⇒ 46 – 7 = 3(n – 1)

⇒ 39 = 3(n – 1)

⇒ n – 1 = 13

⇒ n = 13 + 1

⇒ n = 14

∴ n = 14

Now, we find the sum

⇒

⇒ S_n =7×[53]

⇒ S_n = 371

OR

Given: 9^th term of an AP, a₉ = 0

To show: a₂₉ = 2a₁₉

Proof: We know that,

a_n = a + (n – 1)d

For n = 9

a₉ = a + (9 – 1)d

⇒ 0 = a + 8d [given]

⇒ a = -8d …(i)

Now, For n = 19

a₁₉ = a + (19 – 1)d

⇒ a₁₉ = a + 18d

⇒ a₁₉ = -8d + 18d [from eq. (i)]

⇒ a₁₉ = 10d …(ii)

For n = 29

a₂₉ = a + (29 – 1)d

⇒ a₂₉ = a + 28d

⇒ a₂₉ = -8d + 28d [from eq. (i)]

⇒ a₂₉ = 20d

⇒ a₂₉ = 2 × 10d

⇒ a₂₉ = 2a₁₉ [from eq. (ii)]

Hence Proved