Question

the third term of the AP is 8 and the 9th term exceeds three times the third term by 2. Find the sum of first 19 terms

Answer 1

According to question
Since it is an AP
first term = a
difference = d
nth term = a n

Thus According to question
third term
a (3 ) = a + (n-1) d
      8  = a + (3-1)d
a + 2d = 8           Eqn 1
Similarly
9th term
a (9) = a + 8d      Eqn 2
And According to question
a(9) = 3 a (3) + 2
a + 8d = 3 * 8 + 2
a+ 8d = 26            Ean 3
Thus a = 26 - 8d
Putting a in Eqn 1
26 - 8d + 2d = 8
26 - 8 = 6 d
d= 3
Thus
a = 26 - 24 = 2
Thus
S (19) = 19/2 [ 2*2 + (19-1)3 ]
           = 551

Answer 2

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

So, sum of the first 19 term is 551

Answer.