Question

If ( 2k + 1 ) , ( 4k+ 3 ) , ( 8k + 1 ) are 3 consecutive terms of an A.P, find the
value of K and the 18th term of this sequence.

Answer 1

Given

we have given three consecutive terms of an ap e.g., (2k+1) ,(4k+3) & (8k+1)

To Find

we have to find the value of k and also the 18th term of this sequence.

$\sf\huge\bold{\underline{\underline{{Solution}}}}$

Concept: In an ap the middle term is equal to the average of the two other.

Example: if an ap 1,2,3 [2 is equal to the average of 1 and 3].

Here , first term be (2k+1) second term be (4k+3) and third term be (8k+1)

Then , according to concept

=>2k+1 + 8k+1 ÷2 = 4k+3

=>10k+2/2= 4k+3

=>10k+2= 2(4k+3)

=>10k+2=8k+6

=>10k-8k=6-2

=>2k=4

=>k=4÷2

=>k=2

k=2

First term be 2k+1= 2×2+1=5

second term be =4k+3=4×2+3=11

third term be =8k+1=8×2+1=17

Here ,first term (a) = 5

second term (a₂ )=11

Common difference (d)=a₂ -a= 11-5=6

We have to find 18th term = a+17d

18th term = 5+17×6= 5+102=107

Hence ,value of k=2

18th term of this sequence is 107