Question

The third term of an AP is 8 and the nineth term of an AP exceeds three times the third term by 2. Find the sum of its first 19 terms.
Please solve this question

Answer 1

best of luck for exam ....

Answer 2

Answer:

Step-by-step explanation:

As it is given that third term is 8 so,

a3 = 8

a3 = a+2d

so, a+2d = 8 (let this equation be 1)

As it is given that 9th term exceeds 3 times the 3rd term by 2 so,

As a9 =a+8d so

a+8d - 3(a+2d) = 2

As a+2d = 8 so

a+8d - 3(8) = 2

a+8d - 24 = 2

a+8d =26 (let this equation be 2)

Solve 1 and 2 equations by elimination method

a+2d =8

a+ 8d = 16

- - -

then we get

-4d =-8

d = 2

substitute d = 2 in equation 1

a+2(2) =8

a+4=8

a = 4

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

As n= 19

S19= 19/2(2(4)+(19-1)2

S19= 19/2(8+36)

S19 = 19/2(44)

S19 = 19(22)

S19 = 418

Sum of 19 terms is 418