Question

DATE
1
18. The 3rd term of an AP is 8 & 9th term of AP
exceeds 3 times 3rd term by 2 find the sum of 19 terms​

Answer 1

ANSWER: 551

Given :-

• 3rd term of an ap is 8
• 9th term of the same AP exceeds 3 times 3rd term by 2.

To Find :-

• Sum of first 19 terms.

Solution :-

Given that the 9th term exceeds three times 3rd term by 2 which means,

⇒ 9th term - 3 × 3rd term = 2

Given, 3rd term = 8

⇒ a + 8d - 3a - 6d = 2

⇒ -2a + 2d = 2

⇒ 2a - 2d = -2

⇒ a - d = -1

⇒ a = d - 1 ...(1)

Also, We would need to find the first term of the AP in order to find the sum of first 19 terms. Let's substitute the value of a in 3rd term formula,

⇒ 3rd term = 8

⇒ a + 2d = 8

⇒ d - 1 + 2d = 8 [ from (1) ]

⇒ 3d = 9

⇒ d = 3 ...(2)

Substitute value of d in (1), we get

⇒ a = d - 1

⇒ a = 3 - 1

⇒ a = 2

Now, Let's find the sum,

⇒ Sₙ = n/2 [ 2a + (n - 1)d ]

Here,

• n = 19, d = 3, a = 2

⇒ S₁₉ = 19/2 [ 2×2 + (19 - 1)3 ]

⇒ S₁₉ = 19/2 [ 4 + 18 × 3 ]

⇒ S₁₉ = 19/2 ( 58 )

⇒ S₁₉ = 551

Hence, The sum of first 19 terms is 551.

Answer 2

Answer:

Solution :

Let 'a' be the first term and 'd' be the common difference of the given A.P.

Then, the nth term of the A.P. = an = a + (n - 1)d

Now, it is given that the third term is 8.

Then, a3 = a + (n - 1)d

⇒ a + (3 - 1)d

⇒ a + 2d = 8 ....(1)

Also, the 9th term exceeds 3 times the third term by 2.

Then,

a9 = 3a3 + 2

⇒ a + (9 - 1)d = 3(8) + 2 (As a3 = 8)

⇒ a + 8d = 26 ....(2)

Subtracting (1) from (2)

⇒ a + 8d = 26

a + 2d = 8

(-) (-) (-)

____________

6d = 18

___________

⇒ 6d = 18

⇒ d = 18/6

⇒ d = 3

Putting the value of d = 3 in Equation (1), we get.

⇒ a + 2d = 8

⇒ a + 2 × 3 = 8

⇒ a + 6 = 8

⇒ a = 8 - 6

a = 2

Now, the sum of first n terms of the A.P. is given by

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

Therefore, the sum of first 19 terms of the given A.P. is -

S19 = 19/2[2 × 2 + (19 - 1)3]

⇒ 19/2(4 + 18 × 3)

⇒ 19/2(4 + 54)

⇒ 19/2 × 58

⇒ 1102/2

= 551

Therefore, sum of the first 19 term is 551.