Question

The first three terms of an arithmetic sequence are 2k-7; k+8 and 2k-1.
Calculate the value of the 15th term of the sequence.​

Answer 1

Given:-

• a = 2k - 7
• a₂ = k + 8
• a₃ = 2k - 1

Solution :-

As we know that,

a₃ = a + 2d

So, 2k - 1 = 2k -7 + 2d

2d = 6

d = 3

Again, put value of d in equation,

a₂ = a + d

k + 8 = 2k - 7 + 3

2k - k = 8 + 7 -3

k = 15 - 3

k = 12

⇒ a = 2k - 7

a = 2(12) - 7

a = 24 - 7

a = 17

Now 15th term of sequence will be,

aₙ = a + (n-1)d

a₁₅ = a + (n-1)d

a₁₅ = 17 + (15-1)3

a₁₅ = 17 + (14)3

a₁₅ = 17 + 42

a₁₅ = 59

Therefore, the value of 15th term of the sequence is 59.

Answer 2

Answer:

15th term = 203 - 12k = 59

Step-by-step explanation:

Since it is an AP

Therefore; k+8-(2k-7) = 2k-1-(k+8)

= k+8-2k+7 = 2k-1-k-8

= -k+15 = k-9

= 2k = 15+9 = 24

= k = 12

AP; 2(12)-7, (12)+8, 2(12)-1

17, 20, 23

T15 = a + (n-1) d

= 17 + (15-1) 3

= 17 + 14×3 = 17 + 42

= 59

15 th term = 59